# Operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

Roumaissae Eljazzar1, Mohamed Rossafi2 and Choonkil Park3∗ 1Department of
Mathematics, Faculty of Sciences, University of Ibn Tofail, Kenitra, Morocco
<EMAIL_ADDRESS>2LASMA Laboratory Department of Mathematics
Faculty of Sciences, Dhar El Mahraz University Sidi Mohamed Ben Abdellah, Fes,
Morocco<EMAIL_ADDRESS><EMAIL_ADDRESS>3Research
Institute for Natural Sciences, Hanyang University, Seoul 04763, Korea
<EMAIL_ADDRESS>
# Operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

Roumaissae Eljazzar1, Mohamed Rossafi2 and Choonkil Park3∗ 1Department of
Mathematics, Faculty of Sciences, University of Ibn Tofail, Kenitra, Morocco
<EMAIL_ADDRESS>2LASMA Laboratory Department of Mathematics
Faculty of Sciences, Dhar El Mahraz University Sidi Mohamed Ben Abdellah, Fes,
Morocco<EMAIL_ADDRESS><EMAIL_ADDRESS>3Research
Institute for Natural Sciences, Hanyang University, Seoul 04763, Korea
<EMAIL_ADDRESS>
###### Abstract.

The concept of operator frame can be considered as a generalization of frame.
Firstly, we introduce the notion of operator frame for the set of all
adjointable operators $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ on a Hilbert
pro-$C^{\ast}$-module $\mathcal{X}.$ The analysis operator, the synthesis
operator and the frame operator are presented. Secondly, we study the
stability of operator frame under small perturbations. We also study the
tensor product of operator frame for Hilbert pro-$C^{\ast}$-modules. Finally,
we establish its dual and some properties.

###### Key words and phrases:

Frame; Operator frame; Pro-$C^{\ast}$-algebra; Hilbert pro-$C^{\ast}$-modules;
Tensor product.
∗Corresponding author: Choonkil Park (email<EMAIL_ADDRESS>fax:
+82-2-2281-0019).

###### 2010 Mathematics Subject Classification:

Primary 42C15, 46L05.

##  1\. Introduction

The theory of frames for Hilbert spaces was first introduced in 1952 by Duffin
and Schaefer [4] for the study of nonharmonic series. Later, Frank and Larson
[6, 7] introduced the notion of standard frame of multipliers for a Hilbert
$C^{\ast}$-module and proved some of their basic properties. In 2008, Joita
[9] proposed frames of multipliers in Hilbert pro-$C^{\ast}$-modules and
showed that many properties of frames in Hilbert $C^{\ast}$-modules are valid
for frames of multipliers in Hilbert modules over pro-$C^{\ast}$-algebras.

Operator frames for $B(\mathcal{H})$ is a new notion of frames, which was
introduced by Li and Cao [12] and was more generalized by Rossafi in [16]. For
more information and properties on frames, see [2, 5, 14].

The aim of this work is to introduce the notion of operator frame for the
space $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ of all adjointable operators on
a Hilbert pro-$C^{\ast}$-module for $\mathcal{X}$. We begin by recalling some
fundamental definitions and notations of Hilbert pro-$C^{\ast}$-modules. In
Section 2, we give the definition of operator frame and some properties. In
Section 3, we study the stabilty of operator frame. In Section 4, we
investigate the tensor product of Hilbert pro-$C^{\ast}$-modules and we show
that the tensor product of operator frames for Hilbert pro-$C^{\ast}$-modules
$\mathcal{X}$ and $\mathcal{Y}$ is an operator frame for
$\mathcal{X}\otimes\mathcal{Y}$. Finally, the dual of operator frame and some
properties are discussed.

##  2\. Preliminaries

Recall that a pro-$C^{\ast}$-algebra is a complete Hausdorff complex
topological $\ast$-algebra $\mathcal{A}$ whose topology is determined by its
continuous $C^{\ast}$-seminorms in the sense that a net $\\{a_{\alpha}\\}$
converges to $0$ if and only if $p(a_{\alpha})$ converges to $0$ for all
continuous $C^{\ast}$-seminorms $p$ on $\mathcal{A}$ (see [8, 11, 15]), and we
have

  1. (1)

$p(ab)\leq p(a)p(b)$,

  2. (2)

$p(a^{\ast}a)=p(a)^{2}$

for all $a,b\in\mathcal{A}$.

If the topology of pro-$C^{*}$-algebra is determined by only countably many
$C^{*}$-seminorms, then it is called a $\sigma$-$C^{*}$-algebra.
We denote by $sp(a)$ the spectrum of $a$ such that
$sp(a)=\left\\{\lambda\in\mathbb{C}:\lambda 1_{\mathcal{A}}-a\right.$ is not
invertible$\\}$ for each $a\in\mathcal{A}$, where $\mathcal{A}$ is a unital
pro-$C^{*}$-algebra with unit $1_{\mathcal{A}}$.
The set of all continuous $C^{\ast}$-seminorms on $\mathcal{A}$ is denoted by
$S(\mathcal{A})$. If $\mathcal{A}^{+}$ denotes the set of all positive
elements of $\mathcal{A}$, then $\mathcal{A}^{+}$ is a closed convex
$C^{*}$-seminorms on $\mathcal{A}.$

######  Example 2.1.

Every $C^{*}$-algebra is a pro-$C^{*}$-algebra.

######  Proposition 2.2.

Let $\mathcal{A}$ be a unital pro-$C^{*}$-algebra with unit $1_{\mathcal{A}}.$
Then for any $p\in S(\mathcal{A}),$ we have

  1. (1)

$p(a)=p(a^{*})$ for all $a\in A$;

  2. (2)

$p\left(1_{\mathcal{A}}\right)=1$;

  3. (3)

If $a,b\in\mathcal{A}^{+}$ and $a\leq b$, then $p(a)\leq p(b)$;

  4. (4)

If $1_{\mathcal{A}}\leq b$, then $b$ is invertible and $b^{-1}\leq
1_{\mathcal{A}}$;

  5. (5)

If $a,b\in\mathcal{A}^{+}$ are invertible and $0\leq a\leq b$, then $0\leq
b^{-1}\leq a^{-1}$;

  6. (6)

If $a,b,c\in\mathcal{A}$ and $a\leq b$, then $c^{*}ac\leq c^{*}bc$;

  7. (7)

If $a,b\in\mathcal{A}^{+}$ and $a^{2}\leq b^{2}$, then $0\leq a\leq b$.

######  Definition 2.3.

[15] A pre-Hilbert module over a pro-$C^{\ast}$-algebra $\mathcal{A}$ is a
complex vector space $E$ which is also a left $\mathcal{A}$-module compatible
with the complex algebra structure, equipped with an $\mathcal{A}$-valued
inner product $\langle.,.\rangle$ $E\times E\rightarrow\mathcal{A}$ which is
$\mathcal{A}$-linear in its first variable and satisfies the following
conditions:

  1. 1)

$\langle\xi,\eta\rangle^{*}=\langle\eta,\xi\rangle$;

  2. 2)

$\langle\xi,\xi\rangle\geq 0$;

  3. 3)

$\langle\xi,\xi\rangle=0$ if and only if $\xi=0$

for all $\xi,\eta\in E.$ We say $E$ is a Hilbert $\mathcal{A}$-module (or
Hilbert pro-$C^{\ast}$-module over $\mathcal{A}$). Assume that $E$ is complete
with respect to the topology determined by the family of seminorms

$\bar{p}_{E}(\xi)=\sqrt{p(\langle\xi,\xi\rangle)}\quad\xi\in E,p\in
S(\mathcal{A}).$

Let $\mathcal{A}$ be a pro-$C^{\ast}$-algebra and let $\mathcal{X}$ and
$\mathcal{Y}$ be Hilbert $\mathcal{A}$-modules and assume that $I$ and $J$ are
countable index sets. A bounded $\mathcal{A}$-module map from $\mathcal{X}$ to
$\mathcal{Y}$ is called an operator from $\mathcal{X}$ to $\mathcal{Y}$. We
denote the set of all operators from $\mathcal{X}$ to $\mathcal{Y}$ by
$Hom_{\mathcal{A}}(\mathcal{X},\mathcal{Y})$.

######  Definition 2.4.

An $\mathcal{A}$-module map $T:\mathcal{X}\longrightarrow\mathcal{Y}$ is
adjointable if there is a map $T^{\ast}:\mathcal{Y}\rightarrow\mathcal{X}$
such that $\langle T\xi,\eta\rangle=\left\langle\xi,T^{\ast}\eta\right\rangle$
for all $\xi\in\mathcal{X},\eta\in\mathcal{Y}$, and is called bounded if for
all $p\in S(\mathcal{A})$, there is $M_{p}>0$ such that
$\bar{p}_{\mathcal{Y}}(T\xi)\leq M_{p}\bar{p}_{\mathcal{X}}(\xi)$ for all
$\xi\in\mathcal{X}$.

We denote by $Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$ the set of
all adjointable operators from $\mathcal{X}$ to $\mathcal{Y}$ and let
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})=Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{X})$

######  Definition 2.5.

Let $\mathcal{A}$ be a pro-$C^{\ast}$-algebra and $\mathcal{X},\mathcal{Y}$ be
two Hilbert $\mathcal{A}$-modules. An operator
$T:\mathcal{X}\rightarrow\mathcal{Y}$ is called uniformly bounded below if
there exists $C>0$ such that for each $p\in S(\mathcal{A})$,

$\bar{p}_{\mathcal{Y}}(T\xi)\leqslant C\bar{p}_{\mathcal{X}}(\xi),\quad\text{
for all }\xi\in\mathcal{X}$

and is called uniformly bounded above if there exists $C^{\prime}>0$ such that
for each $p\in S(\mathcal{A})$,

$\bar{p}_{\mathcal{Y}}(T\xi)\geqslant
C^{\prime}\bar{p}_{\mathcal{X}}(\xi),\quad\text{ for all }\xi\in\mathcal{X}.$

Let

$\|T\|_{\infty}=\inf\\{M:M\text{ is an upper bound for }T\\},$
$\hat{p}_{\mathcal{Y}}(T)=\sup\left\\{\bar{p}_{\mathcal{Y}}(T(x)):\xi\in\mathcal{X},\quad\bar{p}_{\mathcal{X}}(\xi)\leqslant
1\right\\}.$

It is clear to see that $\hat{p}(T)\leqslant\|T\|_{\infty}$ for all $p\in
S(\mathcal{A})$.

######  Proposition 2.6.

[3] Let $\mathcal{X}$ be a Hilbert module over pro-$C^{*}$-algebra
$\mathcal{A}$ and $T$ be an invertible element in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$, which is uniformly bounded. Then for
all $\xi\in\mathcal{X}$,

$\left\|T^{-1}\right\|_{\infty}^{-2}\langle\xi,\xi\rangle\leq\langle
T\xi,T\xi\rangle\leq\|T\|_{\infty}^{2}\langle\xi,\xi\rangle.$

######  Definition 2.7.

A sequence $\\{x_{i}\\}_{i\in J}$ of elements in a Hilbert
$\mathcal{A}$-module $\mathcal{X}$ is said to be a frame if there are real
constants $C,D>0$ such that:

$C\langle\xi,\xi\rangle\leq\sum_{i\in
J}\left\langle\xi,x_{i}\right\rangle\left\langle x_{i},\xi\right\rangle\leq
D\langle\xi,\xi\rangle$ (2.1)

for all $\xi\in\mathcal{X}$. The constants $C$ and $D$ are called lower and
upper frame bounds for the frame, respectively. If $C=D=\lambda$, the frame is
called a $\lambda$-tight frame. If $C=D=1$, the frame is called a Parseval
frame. We say that $\\{x_{i}\\}_{i\in J}$ is a frame sequence if it is a frame
for the closure of $\mathcal{A}$-linear hull of $\\{x_{i}\\}_{i\in J}$. If, in
(2.1), we only require to have the upper bound, then $\\{x_{i}\\}_{i\in J}$ is
called a Bessel sequence with Bessel bound $D$.

######  Definition 2.8.

[3] Let $\\{v_{i}\\}_{i\in J}$ be a family of weights in $A$, i.e., each
$v_{i}$ is a positive invertible element from the center of $\mathcal{A}$, and
let $\left\\{M_{i}:i\in J\right\\}$ be a family of orthogonally complemented
submodules of $\mathcal{X}.$ Then $\left\\{\left(M_{i},v_{i}\right):i\in
J\right\\}$ is a fusion frame if there exist real constants $C,D>$ 0 such
that:

$C\langle\xi,\xi\rangle\leq\sum_{i\in J}v_{i}^{2}\left\langle
P_{M_{i}}(\xi),P_{M_{i}}(\xi)\right\rangle\leq D\langle\xi,\xi\rangle$ (2.2)

for all $\xi\in\mathcal{X}$. We call $C$ and $D$ the lower and upper bounds of
the fusion frame. If $C=D=\lambda$, then the family
$\left\\{\left(M_{i},v_{i}\right):i\in J\right\\}$ is called a $\lambda$-tight
fusion frame and if $C=D=1$, then it is called a Parseval fusion frame. If, in
(2.2), we only have the upper bound, then
$\left\\{\left(M_{i},v_{i}\right):i\in J\right\\}$ is called a Bessel fusion
sequence with Bessel bound $D$.

See [10, 13] for more infornation on fusion frames and applications.

######  Lemma 2.9.

[1] Let $\mathcal{A}$ be a locally $C^{\ast}$-algebra, $\mathcal{X}$ be a
Hilbert $\mathcal{A}$-module and $T$ be a bounded operator in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ such that $T^{*}=T.$ The following
statements are mutually equivalent:

  1. (1)

$T$ is surjective.

  2. (2)

There are $m,M>0$ such that
$m\bar{p}_{\mathcal{X}}(\xi)\leq\bar{p}_{\mathcal{X}}(T\xi)\leq
M\bar{p}_{\mathcal{X}}(\xi)$ for all $p\in S(\mathcal{A})$ and
$\xi\in\mathcal{X}$.

  3. (3)

There are $m^{\prime},M^{\prime}>0$ such that
$m^{\prime}\langle\xi,\xi\rangle\leq\langle T\xi,T\xi\rangle\leq
M^{\prime}\langle\xi,\xi\rangle$ for all $\xi\in\mathcal{X}.$

Similar to $C^{\ast}$-algebra, each $\ast$-homomorphism between two
pro-$C^{\ast}$-algebra is increasing.

######  Lemma 2.10.

If $\varphi:\mathcal{A}\rightarrow\mathcal{B}$ is a $\ast$-homomorphism
between pro-$\mathcal{C}^{\ast}$-algebras, then $\varphi$ is increasing, that
is, if $a\leq b$, then $\varphi(a)\leq\varphi(b)$.

###### Proof.

The proof is easy. ∎

##  3\. Operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

######  Definition 3.1.

A family of adjointable operators $\\{T_{i}\\}_{i\in J}$ on a Hilbert
$\mathcal{A}$-module $\mathcal{X}$ over a unital pro-$C^{\ast}$-algebra is
said to be an operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ if
there exist constants $A,B>0$ such that

$A\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle T_{i}\xi,T_{i}\xi\rangle\leq
B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{X}.$ (3.1)

The numbers $A$ and $B$ are called lower and upper bounds of the operator
frame, respectively. If $A=B=\lambda$, then the operator frame is
$\lambda$-tight. If $A=B=1$, then it is called a normalized tight operator
frame or a Parseval operator frame. If only upper inequality of (3.1) holds,
then $\\{T_{i}\\}_{i\in J}$ is called an operator Bessel sequence for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.

If the sum in the middle of (3.1) is convergent in norm, then the operator
frame is called standard.

######  Example 3.2.

Let $\mathcal{A}$ be a Hilbert pro-$C^{\ast}$-module over itself with the
inner product $\langle a,b\rangle=ab^{\ast}$. Let $\\{x_{i}\\}_{i\in J}$ be a
frame for $\mathcal{A}$ with bounds $A$ and $B$, respectively. For each $i\in
J$, we define $T_{i}:\mathcal{A}\to\mathcal{A}$ by
$T_{i}\xi=\langle\xi,x_{i}\rangle,\;\;\forall\xi\in\mathcal{A}$. Then $T_{i}$
is adjointable and $T_{i}^{\ast}a=ax_{i}$ for all $a\in\mathcal{A}$. We have

$A\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle\xi,x_{i}\rangle\langle
x_{i},\xi\rangle\leq B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{A}$

and so

$A\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle T_{i}\xi,T_{i}\xi\rangle\leq
B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{A}.$

Thus $\\{T_{i}\\}_{i\in J}$ is an operator frame in $\mathcal{A}$ with bounds
$A$ and $B$, respectively.

######  Example 3.3.

Let $\\{W_{i}\\}_{i\in J}$ be a frame of submodules with respect to
$\\{v_{i}\\}_{i\in J}$ for $\mathcal{X}$. Put $T_{i}=v_{i}\pi_{W_{i}},\forall
i\in J$. Then we get a sequence of operators $\\{T_{i}\\}_{i\in J}$ and there
exist constants $A,B>0$ such that

$A\langle\xi,\xi\rangle\leq\sum_{i\in J}v_{i}^{2}\langle
P_{W_{i}}\xi,P_{W_{i}}\xi\rangle\leq
B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{X}.$

So we have

$A\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle T_{i}\xi,T_{i}\xi\rangle\leq
B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{X}.$

Thus the sequence $\\{T_{i}\\}_{i\in J}$ becomes an operator frame for
$\mathcal{X}$.

With this example, a frame of submodules can be viewed as a special case of
operator frames.

######  Theorem 3.4.

Let $\\{T_{i}\\}_{i\in J}$ be an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ with bounds $A$ and $B$. Let $Q$ be a
bounded element in $Hom_{\mathcal{A}}(\mathcal{X})$ such that $Q=Q^{\ast}.$ If
$Q$ is surjective, then $\\{T_{i}Q\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X}).$

###### Proof.

Since $Q$ is surjective, by Lemma 2.9, there are $m,M>0$ such that

$m\langle\xi,\xi\rangle\leq\langle Q\xi,Q\xi\rangle\leq
M\langle\xi,\xi\rangle\quad\text{for all}\quad\xi\in\mathcal{X}.$

Then

$Am\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle
T_{i}Q\xi,T_{i}Q\xi\rangle\leq
BM\langle\xi,\xi\rangle,\forall\xi\in\mathcal{X}.$

Hence $\\{T_{i}Q\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X}).$ ∎

Let $\\{T_{i}\\}_{i\in J}$ be an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. Define an operator

$R:\mathcal{X}\rightarrow l^{2}(\mathcal{X})\;\;by\;\;Rx=\\{T_{i}x\\}_{i\in
J},\forall x\in\mathcal{X}.$

The operator $R$ is called an analysis operator of the operator frame
$J\\{T_{i}\\}_{i\in J}$.
The adjoint $R^{\ast}(\\{x_{i}\\}_{i\in
J}):l^{2}(\mathcal{X})\rightarrow\mathcal{X}$ of the analysis operator $R$ is
defined by

$R^{\ast}(\\{x_{i}\\}_{i\in J})=\sum_{i\in
J}T_{i}^{\ast}x_{i},\forall\\{x_{i}\\}_{i\in J}\in l^{2}(\mathcal{X}).$

The operator $R^{\ast}$ is called the synthesis operator of the operator frame
$\\{T_{i}\\}_{i\in J}$.
By composing $R$ and $R^{\ast}$, the frame operator
$S:\mathcal{X}\rightarrow\mathcal{X}$ for the operator frame is given by

$S(x)=R^{\ast}Rx=\sum_{i\in J}T_{i}^{\ast}T_{i}x.$

######  Theorem 3.5.

Let $\\{T_{i}\\}_{i\in J}$ be a family of adjointable operators on a Hilbert
$\mathcal{A}$-module $\mathcal{X}$. Then $\\{T_{i}\\}_{i\in J}$ is an operator
frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ if and only if there exist
positive constants $A,B>0$ such that

$A\bar{p}_{\mathcal{X}}(\xi)^{2}\leq p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)\leq
B\bar{p}_{\mathcal{X}}(\xi)^{2},\forall\xi\in\mathcal{X}.$ (3.2)

###### Proof.

Suppose that $\\{T_{i}\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. Combined with the definition of an
operator frame, we know that (3.2) holds.

Now suppose that (3.2) holds. We know that the frame operator $S$ is positive,
self-adjoint and invertible, and hence

$\langle S^{\frac{1}{2}}x,S^{\frac{1}{2}}\xi\rangle=\langle
S\xi,\xi\rangle=\sum_{i\in J}\langle T_{i}\xi,T_{i}\xi\rangle.$

So we have $Ap(\langle\xi,\xi\rangle)\leq p(\langle
S^{\frac{1}{2}}\xi,S^{\frac{1}{2}}\xi\rangle)\leq Bp(\langle\xi,\xi\rangle)$
for all $\xi\in\mathcal{X}.$ Thus
$A\bar{p}_{\mathcal{X}}(\xi)^{2}\leq\bar{p}_{\mathcal{Y}}(S^{\frac{1}{2}}\xi)^{2}\leq
B\bar{p}_{\mathcal{X}}(\xi)^{2}.$ According to Proposition 2.6, we have

$\|S^{-\frac{1}{2}}\|_{\infty}^{-2}\langle\xi,\xi\rangle\leq\langle
S^{\frac{1}{2}}\xi,S^{\frac{1}{2}}\xi\rangle=\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle\leq\|S^{\frac{1}{2}}\|_{\infty}^{2}\langle\xi,\xi\rangle,$

which implies that $\\{T_{i}\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. ∎

######  Theorem 3.6.

Assume that $S$ is the frame operator of an operator frame
$T=\\{T_{i}\\}_{i\in J}$ for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ with
bounds $A$ and $B$. Then $S$ is positive, self-adjoint and invertible.
Moreover, we have $AI\leq S\leq BI$ and the reconstruction formula

$\xi=\sum_{i\in J}S^{-1}T_{i}^{\ast}T_{i}\xi,\forall\xi\in\mathcal{X}.$ (3.3)

###### Proof.

It is clear that $S$ is positive and self-adjoint. For any
$\xi\in\mathcal{X}$, since $T=\\{T_{i}\\}_{i\in J}$ is an operator frame with
bounds $A,B$, we have

$\langle A\xi,\xi\rangle=A\langle\xi,\xi\rangle\leq\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle=\langle S\xi,\xi\rangle\leq
B\langle\xi,\xi\rangle=\langle B\xi,\xi\rangle.$

This shows that

$AI\leq S\leq BI,$

which implies that $S$ is invertible. Further, for any $\xi\in\mathcal{X}$, we
have

$\xi=S_{T}^{-1}S_{T}\xi=S_{T}^{-1}\sum_{i\in J}T_{i}^{\ast}T_{i}\xi=\sum_{i\in
J}S_{T}^{-1}T_{i}^{\ast}T_{i}\xi.$

This completes the proof. ∎

######  Theorem 3.7.

Let $(\mathcal{X},\mathcal{A},\langle.,.\rangle_{\mathcal{A}})$ and
$(\mathcal{X},\mathcal{B},\langle.,.\rangle_{\mathcal{B}})$ be two Hilbert
pro-$\mathcal{C^{\ast}}$-modules and
$\varphi:\mathcal{A}\longrightarrow\mathcal{B}$ be a $\ast$-homomorphism and
$\theta\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be an invertible map such
that both are uniformly bounded and
$\langle\theta\xi,\theta\eta\rangle_{\mathcal{B}}=\varphi(\langle\xi,\eta\rangle_{\mathcal{A}})$
for all $\xi,\eta\in\mathcal{X}$. Suppose that $\\{T_{i}\\}_{i\in J}$ is an
operator frame for $(\mathcal{X},\mathcal{A},\langle.,.\rangle_{\mathcal{A}})$
with frame operator $S_{\mathcal{A}}$ and lower and upper operator frame
bounds $A$ and $B$, respectively. Then $\\{\theta T_{i}\\}_{i}$ is an operator
frame for $(\mathcal{X},\mathcal{B},\langle.,.\rangle_{\mathcal{B}})$ with
frame operator $S_{\mathcal{B}}$ and lower and upper operator frame bounds
$\|\theta^{-1}\|_{\infty}^{-2}A$ ,$\|\theta\|_{\infty}^{2}B$, respectively,
and $\langle S_{\mathcal{B}}\xi,\eta\rangle_{\mathcal{B}}=\varphi(\langle
S_{\mathcal{A}}\xi,\eta\rangle_{\mathcal{A}})$.

###### Proof.

By the definition of operator frame, we have

$A\langle\xi,\xi\rangle_{\mathcal{A}}\leq\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle_{\mathcal{A}}\leq
B\langle\xi,\xi\rangle_{\mathcal{A}},\forall\xi\in\mathcal{X}.$

By Lemma 2.10, we have

$\varphi(A\langle\xi,\xi\rangle_{\mathcal{A}})\leq\varphi(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle_{\mathcal{A}})\leq\varphi(B\langle\xi,\xi\rangle_{\mathcal{A}}),\forall\xi\in\mathcal{X}.$

By the definition of $\ast$-homomorphism, we have

$A\varphi(\langle\xi,\xi\rangle_{\mathcal{A}})\leq\sum_{i\in J}\varphi(\langle
T_{i}\xi,T_{i}\xi\rangle_{\mathcal{A}})\leq
B\varphi(\langle\xi,\xi\rangle_{\mathcal{A}}),\forall\xi\in\mathcal{X}.$

By the relation betwen $\theta$ and $\varphi$, we get

$A\langle\theta\xi,\theta\xi\rangle_{\mathcal{B}}\leq\sum_{i\in
J}\langle\theta T_{i}\xi,\theta T_{i}\xi\rangle_{\mathcal{B}}\leq
B\langle\theta\xi,\theta\xi\rangle_{\mathcal{B}},\forall\xi\in\mathcal{X}.$

By Proposition 2.6, we have

$\|\theta^{-1}\|_{\infty}^{-2}\langle\xi,\xi\rangle_{\mathcal{B}}\leq\langle\theta\xi,\theta\xi\rangle_{\mathcal{B}}\leq\|\theta\|_{\infty}^{2}\langle\xi,\xi\rangle_{\mathcal{B}}.$

Thus

$\|\theta^{-1}\|_{\infty}^{-2}A\langle\xi,\xi\rangle_{\mathcal{B}}\leq\sum_{i\in
J}\langle\theta T_{i}\xi,\theta
T_{i}\xi\rangle_{\mathcal{B}}\leq\|\theta\|_{\infty}^{2}B\langle\xi,\xi\rangle_{\mathcal{B}},\quad\forall\xi\in\mathcal{X}.$

On the other hand, we have

$\displaystyle\varphi(\langle S_{\mathcal{A}}\xi,\eta\rangle_{\mathcal{A}})$
$\displaystyle=\varphi(\langle\sum_{i\in
J}T_{i}^{\ast}T_{i}\xi,\eta\rangle_{\mathcal{A}})$ $\displaystyle=\sum_{i\in
J}\varphi(\langle T_{i}\xi,T_{i}\eta\rangle_{\mathcal{A}})$
$\displaystyle=\sum_{i\in J}\langle\theta T_{i}\xi,\theta
T_{i}\eta\rangle_{\mathcal{B}}$ $\displaystyle=\sum_{i\in J}\langle(\theta
T_{i})^{\ast}(\theta T_{i})\xi,\eta\rangle_{\mathcal{B}}$
$\displaystyle=\langle\sum_{i\in J}(\theta T_{i})^{\ast}(\theta
T_{i})\xi,\eta\rangle_{\mathcal{B}}$ $\displaystyle=\langle
S_{\mathcal{B}}\xi,\eta\rangle_{\mathcal{B}}.$

This completes the proof. ∎

##  4\. Perturbation and stability of operator frames for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

######  Theorem 4.1.

Let $\left\\{T_{i}\right\\}_{i\in J}$ be an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ with bounds $A$ and B. If
$\left\\{R_{i}\right\\}_{i\in J}\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ is
an operator Bessel sequence with a bound $M<A$, then
$\left\\{T_{i}-R_{i}\right\\}_{i\in\mathbb{J}}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.

###### Proof.

For any $\xi\in\mathcal{X}$, we have

$\displaystyle p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)^{\frac{1}{2}}$
$\displaystyle=\bar{p}_{\mathcal{X}}\\{(T_{i}-R_{i})\xi\\}_{i\in J}$
$\displaystyle\leq p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}+p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle\leq\sqrt{B}\bar{p}_{\mathcal{X}}(\xi)+\sqrt{M}\bar{p}_{\mathcal{X}}(\xi)$
$\displaystyle\leq(\sqrt{B}+\sqrt{M})\bar{p}_{\mathcal{X}}(\xi).$

On the other hand,

$\displaystyle Ap(\langle\xi,\xi\rangle)$ $\displaystyle\leq p(\sum_{i\in
J}\langle T_{i}\xi,T_{i}\xi\rangle)$ $\displaystyle\leq p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)+p(\sum_{i\in
J}\langle R_{i}\xi,R_{i}\xi\rangle)$ $\displaystyle\leq p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)+Mp(\langle\xi,\xi\rangle).$

Thus

$(A-M)p(\langle\xi,\xi\rangle)\leq p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle).$

Consequently,

$(A-M)^{2}\bar{p}_{\mathcal{X}}(\xi)^{2}\leq p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)\leq(\sqrt{B}+\sqrt{M})^{2}\bar{p}_{\mathcal{X}}(\xi)^{2}.$

Hence $\left\\{T_{i}-R_{i}\right\\}_{i\in\mathbb{J}}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. ∎

######  Theorem 4.2.

Let $\\{T_{i}\\}_{i\in J}$ be an operator frame for
$Hom_{\mathcal{A}}^{*}(\mathcal{X})$ with bound $A$ and $B$ and
$\left\\{R_{i}\right\\}_{i\in J}\in Hom_{\mathcal{A}}^{*}(\mathcal{X}).$ Then
the following statements are equivalent:

  1. (i)

$\left\\{R_{i}\right\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{*}(\mathcal{X}).$

  2. (ii)

There exists a constant $M>0$ such that for all $x\in\mathcal{X}$, we have

$\displaystyle p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)$
(4.1) $\displaystyle\qquad\leq M\min\left(p(\sum_{i\in J}\left\langle
T_{i}\xi,T_{i}\xi\right\rangle),p(\sum_{i\in J}\left\langle
R_{i}\xi,R_{i}\xi\right\rangle)\right).$

###### Proof.

First, suppose that $\left\\{R_{i}\right\\}_{i\in J}\in
Hom_{\mathcal{A}}^{*}(\mathcal{X})$ is an operator frame with bounds $C$ and
$D$. Then for any $\xi\in\mathcal{X}$, we have

$\displaystyle p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)^{\frac{1}{2}}$
$\displaystyle=\bar{p}_{\mathcal{X}}\\{(T_{i}-R_{i})\xi\\}_{i\in J}$
$\displaystyle\leq p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}+p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}$ $\displaystyle\leq p(\sum_{i\in
J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}+\sqrt{D}\bar{p}_{\mathcal{X}}(\xi)$
$\displaystyle\leq p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}+\sqrt{\frac{D}{A}}p(\sum_{i\in
J}\langle T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle\leq(1+\sqrt{\frac{D}{A}})p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}.$

Similarly, we can obtain

$p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)^{\frac{1}{2}}\leq(1+\sqrt{\frac{B}{C}})p(\sum_{i\in
J}\langle R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}.$

Let $M=\min\left\\{1+\sqrt{\frac{D}{A}},1+\sqrt{\frac{B}{C}}\right\\}$. Then
(4.1) holds.

Conversely, suppose that (4.1) holds. For every $\xi\in\mathcal{X}$, we have

$\displaystyle\sqrt{A}\bar{p}_{\mathcal{X}}(\xi)$ $\displaystyle\leq
p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}=\bar{p}_{\mathcal{X}}\\{T_{i}\xi\\}_{i\in
J}$ $\displaystyle\leq\bar{p}_{\mathcal{X}}\\{(T_{i}-R_{i})\xi\\}_{i\in
J}+\bar{p}_{\mathcal{X}}\\{R_{i}\xi\\}_{i\in J}$ $\displaystyle\leq
p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)^{\frac{1}{2}}+p(\sum_{i\in
J}\langle R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle\leq\sqrt{M}p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}+p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle=(\sqrt{M}+1)p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}.$

Also we have

$\displaystyle p(\sum_{i\in J}\langle
R_{i}\xi,R_{i}\xi\rangle)^{\frac{1}{2}}=\bar{p}_{\mathcal{X}}\\{R_{i}\xi\\}_{i\in
J}$ $\displaystyle\leq\bar{p}_{\mathcal{X}}\\{(T_{i}-R_{i})\xi\\}_{i\in
J}+\bar{p}_{\mathcal{X}}\\{T_{i}\xi\\}_{i\in J}$ $\displaystyle\leq
p(\sum_{i\in
J}\left\langle\left(T_{i}-R_{i}\right)\xi,\left(T_{i}-R_{i}\right)\xi\right\rangle)^{\frac{1}{2}}+p(\sum_{i\in
J}\langle T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle\leq\sqrt{M}p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}+p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}$
$\displaystyle=(\sqrt{M}+1)p(\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle)^{\frac{1}{2}}.$

Thus

$\frac{A}{(\sqrt{M}+1)^{2}}\bar{p}_{\mathcal{X}}(\xi)^{2}\leq
p(\sum_{i\in\mathbb{J}}\left\langle R_{i}x,R_{i}x\right\rangle)\leq
B(\sqrt{M}+1)^{2}\bar{p}_{\mathcal{X}}(\xi)^{2}.$

So $\left\\{R_{i}\right\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{*}(\mathcal{X}).$ ∎

##  5\. Tensor product

The minimal or injective tensor product of the pro-$C^{\ast}$-algebras
$\mathcal{A}$ and $\mathcal{B}$, denoted by $\mathcal{A}\otimes\mathcal{B}$,
is the completion of the algebraic tensor product
$\mathcal{A}\otimes_{\text{alg }}\mathcal{B}$ with respect to the topology
determined by a family of $C^{\ast}$-seminorms. Suppose that $\mathcal{X}$ is
a Hilbert module over a pro-$C^{\ast}$-algebra $\mathcal{A}$ and $\mathcal{Y}$
is a Hilbert module over a pro-$C^{\ast}$-algebra $\mathcal{B}$. The algebraic
tensor product $\mathcal{X}\otimes_{\text{alg }}\mathcal{Y}$ of $\mathcal{X}$
and $\mathcal{Y}$ is a pre-Hilbert $\mathcal{A}\otimes\mathcal{B}$-module with
the action of $\mathcal{A}\otimes\mathcal{B}$ on
$\mathcal{X}\otimes_{\text{alg }}\mathcal{Y}$ defined by

$(\xi\otimes\eta)(a\otimes b)=\xi a\otimes\eta b$

for all $\xi\in\mathcal{X},\eta\in\mathcal{Y},a\in\mathcal{A}$ and
$b\in\mathcal{B}$ and the inner product

$\langle\cdot,\cdot\rangle:\left(\mathcal{X}\otimes_{\text{alg
}}\mathcal{Y}\right)\times\left(\mathcal{X}\otimes_{\text{alg
}}\mathcal{Y}\right)\rightarrow\mathcal{A}\otimes_{\text{alg }}\mathcal{B}$

is defined by

$\left\langle\xi_{1}\otimes\eta_{1},\xi_{2}\otimes\eta_{2}\right\rangle=\left\langle\xi_{1},\xi_{2}\right\rangle\otimes\left\langle\eta_{1},\eta_{2}\right\rangle.$

We also know that for $z=\sum_{i=1}^{n}\xi_{i}\otimes\eta_{i}$ in
$\mathcal{X}\otimes_{alg}\mathcal{Y}$, we have $\langle
z,z\rangle_{\mathcal{A}\otimes\mathcal{B}}=\sum_{i,j}\langle\xi_{i},\xi_{j}\rangle_{\mathcal{A}}\otimes\langle\eta_{i},\eta_{j}\rangle_{\mathcal{B}}\geq
0$ and $\langle z,z\rangle_{\mathcal{A}\otimes\mathcal{B}}=0$ if and only if
$z=0$.
The external tensor product of $\mathcal{X}$ and $\mathcal{Y}$ is the Hilbert
module $\mathcal{X}\otimes\mathcal{Y}$ over $\mathcal{A}\otimes\mathcal{B}$
obtained by the completion of the pre-Hilbert
$\mathcal{A}\otimes\mathcal{B}$-module $\mathcal{X}\otimes_{\text{alg
}}\mathcal{Y}$.

If $P\in M(\mathcal{X})$ and $Q\in M(\mathcal{Y})$ then there is a unique
adjointable module morphism $P\otimes
Q:\mathcal{A}\otimes\mathcal{B}\rightarrow\mathcal{X}\otimes\mathcal{Y}$ such
that $(P\otimes Q)(a\otimes b)=P(a)\otimes Q(b)$ and $(P\otimes
Q)^{*}(a\otimes b)=P^{*}(a)\otimes Q^{*}(b)$ for all $a\in A$ and all $b\in B$
(see [9]).

######  Theorem 5.1.

Let $\mathcal{X}$ and $\mathcal{Y}$ be two Hilbert pro-$C^{\ast}$-modules over
unitary pro-$C^{\ast}$-algebras $\mathcal{A}$ and $\mathcal{B}$, respectively.
Let $\\{T_{i}\\}_{i\in I}$ and $\\{L_{j}\\}_{j\in J}$ be two operator frames
for $\mathcal{X}$ and $\mathcal{Y}$ with frame operators $S_{T}$ and $S_{L}$
and operator frame bounds $(A,B)$ and $(C,D)$, respectively. Then
$\\{T_{i}\otimes L_{j}\\}_{i\in I,j\in J}$ is an operator frame for the Hibert
$\mathcal{A}\otimes\mathcal{B}$-module $\mathcal{X}\otimes\mathcal{Y}$ with
frame operator $S_{T}\otimes S_{L}$ and lower and upper operator frame bounds
$A\otimes C$ and $B\otimes D$, respectively.

###### Proof.

By the definition of operator frames $\\{T_{i}\\}_{i\in I}$ and
$\\{L_{j}\\}_{j\in J}$, we have

$A\langle\xi,\xi\rangle\leq\sum_{i\in I}\langle T_{i}\xi,T_{i}\xi\rangle\leq
B\langle\xi,\xi\rangle,\forall\xi\in\mathcal{X},$
$C\langle\eta,\eta\rangle\leq\sum_{j\in J}\langle
L_{j}\eta,L_{j}\eta\rangle\leq
D\langle\eta,\eta\rangle,\forall\eta\in\mathcal{Y}.$

Thus

$\displaystyle(A\langle\xi,\xi\rangle)\otimes(C\langle\eta,\eta\rangle)$
$\displaystyle\leq\sum_{i\in I}\langle
T_{i}\xi,T_{i}\xi\rangle\otimes\sum_{j\in J}\langle
L_{j}\eta,L_{j}\eta\rangle$
$\displaystyle\leq(B\langle\xi,\xi\rangle)\otimes(D\langle\eta,\eta\rangle),\forall\xi\in\mathcal{X},\forall\eta\in\mathcal{Y}.$

So

$\displaystyle(A\otimes
C)(\langle\xi,\xi\rangle\otimes\langle\eta,\eta\rangle)$
$\displaystyle\leq\sum_{i\in I,j\in J}\langle
T_{i}\xi,T_{i}\xi\rangle\otimes\langle L_{j}\eta,L_{j}\eta\rangle$
$\displaystyle\leq(B\otimes
D)(\langle\xi,\xi\rangle\otimes\langle\eta,\eta\rangle),\forall\xi\in\mathcal{X},\forall\eta\in\mathcal{Y}.$

Consequently, we have

$\displaystyle(A\otimes C)\langle\xi\otimes\eta,\xi\otimes\eta\rangle$
$\displaystyle\leq\sum_{i\in I,j\in J}\langle T_{i}\xi\otimes
L_{j}\eta,T_{i}\xi\otimes L_{j}\eta\rangle$ $\displaystyle\leq(B\otimes
D)\langle\xi\otimes\eta,\xi\otimes\eta\rangle,\forall\xi\in\mathcal{X},\forall\eta\in\mathcal{Y}.$

Hence for all $\xi\otimes\eta\in\mathcal{X\otimes Y}$, we have

$\displaystyle(A\otimes C)\langle\xi\otimes\eta,\xi\otimes\eta\rangle$
$\displaystyle\leq\sum_{i\in I,j\in J}\langle(T_{i}\otimes
L_{j})(\xi\otimes\eta),(T_{i}\otimes L_{j})(\xi\otimes\eta)\rangle$
$\displaystyle\leq(B\otimes D)\langle\xi\otimes\eta,\xi\otimes\eta\rangle.$

The last inequality is satisfied for every finite sum of elements in
$\mathcal{X}\otimes_{alg}\mathcal{Y}$ and so it is satisfied for all
$z\in\mathcal{X\otimes Y}$. This shows that $\\{T_{i}\otimes L_{j}\\}_{i\in
I,j\in J}$ is an operator frame for the Hibert
$\mathcal{A}\otimes\mathcal{B}$-module $\mathcal{X}\otimes\mathcal{Y}$ with
lower and upper operator frame bounds $A\otimes C$ and $B\otimes D$,
respectively.

By the definition of frame operators $S_{T}$ and $S_{L}$, we have

$S_{T}\xi=\sum_{i\in I}T_{i}^{\ast}T_{i}\xi,\quad\forall\xi\in\mathcal{X},$
$S_{L}\eta=\sum_{j\in J}L_{j}^{\ast}L_{j}\eta,\quad\forall\eta\in\mathcal{Y}.$

Thus

$\displaystyle(S_{T}\otimes S_{L})(\xi\otimes\eta)$
$\displaystyle=S_{T}\xi\otimes S_{L}\eta$ $\displaystyle=\sum_{i\in
I}T_{i}^{\ast}T_{i}\xi\otimes\sum_{j\in J}L_{j}^{\ast}L_{j}\eta$
$\displaystyle=\sum_{i\in I,j\in J}T_{i}^{\ast}T_{i}\xi\otimes
L_{j}^{\ast}L_{j}\eta$ $\displaystyle=\sum_{i\in I,j\in J}(T_{i}^{\ast}\otimes
L_{j}^{\ast})(T_{i}\xi\otimes L_{j}\eta)$ $\displaystyle=\sum_{i\in I,j\in
J}(T_{i}^{\ast}\otimes L_{j}^{\ast})(T_{i}\otimes L_{j})(\xi\otimes\eta)$
$\displaystyle=\sum_{i\in I,j\in J}(T_{i}\otimes L_{j})^{\ast}(T_{i}\otimes
L_{j})(\xi\otimes\eta).$

Now by the uniqueness of frame operator, the last expression is equal to
$S_{T\otimes L}(\xi\otimes\eta)$. Consequently, we have $(S_{T}\otimes
S_{L})(\xi\otimes\eta)=S_{T\otimes L}(\xi\otimes\eta)$. The last equality is
satisfied for every finite sum of elements in
$\mathcal{X}\otimes_{alg}\mathcal{Y}$ and so it is satisfied for all
$z\in\mathcal{X\otimes Y}$. This shows that $(S_{T}\otimes
S_{L})(z)=S_{T\otimes L}(z)$. So $S_{T\otimes L}=S_{T}\otimes S_{L}$. ∎

##  6\. Dual of operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

######  Definition 6.1.

Let $T=\\{T_{i}\\}_{i\in J}$ be an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. A family of operators
$\tilde{T}=\\{\tilde{T}_{i}\\}_{i\in J}$ on $\mathcal{X}$ is called a dual of
the operator frame $\\{T_{i}\\}_{i\in J}$ if they satisfy

$\xi=\sum_{i\in J}T_{i}^{\ast}\tilde{T}_{i}\xi,\quad\forall\xi\in\mathcal{X}.$
(6.1)

Furthermore, we call $\\{\tilde{T}_{i}\\}_{i\in J}$ a dual frame of the
operator frame $\\{T_{i}\\}_{i\in J}$ if $\\{\tilde{T}_{i}\\}_{i\in J}$ is
also an operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and
satisfies (6.1).

######  Theorem 6.2.

Every operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ has a dual
frame.

###### Proof.

If $T=\\{T_{i}\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ with bounds $A$ and $B$, then the
operator sequence $\tilde{T}=\\{T_{i}S_{T}^{-1}\\}_{i\in J}$ is a dual frame
of $T=\\{T_{i}\\}_{i\in J}$. In fact, we have

$\xi=S_{T}S_{T}^{-1}\xi=\sum_{i\in J}T_{i}^{\ast}T_{i}S_{T}^{-1}\xi=\sum_{i\in
J}T_{i}^{\ast}\tilde{T}_{i}\xi,\quad\forall\xi\in\mathcal{X}.$

And $\tilde{T}=\\{T_{i}S_{T}^{-1}\\}_{i\in J}$ satisfies

$A\|S_{T}\|_{\infty}^{-2}\langle\xi,\xi\rangle\leq\sum_{i\in
J}\langle\tilde{T}_{i}\xi,\tilde{T}_{i}\xi\rangle=\sum_{i\in
J}\langle\tilde{T}_{i}S_{T}^{-1}\xi,\tilde{T}_{i}S_{T}^{-1}\xi\rangle\leq
B\|S_{T}^{-1}\|_{\infty}^{2}\langle\xi,\xi\rangle$

for all $\xi\in\mathcal{X}$. Hence $\\{T_{i}S_{T}^{-1}\\}_{i\in J}$ is a dual
frame of $\\{T_{i}\\}_{i\in J}$., which is called the canonical dual frame of
$\\{T_{i}\\}_{i\in J}$. ∎

######  Remark 6.3.

Assume that $T=\\{T_{i}\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ with analytic operator $R_{T}$ and
$\tilde{T}=\\{\tilde{T}_{i}\\}_{i\in J}$ is a dual frame of $T$ with analytic
operator $R_{\tilde{T}}$. Then for any $\xi$ in $\mathcal{X}$, we have

$\xi=\sum_{i\in J}T_{i}^{\ast}\tilde{T}_{i}\xi=R_{T}^{\ast}R_{\tilde{T}}\xi.$

This shows that every element of $\mathcal{X}$ can be reconstructed with an
operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and its dual frame.
Moreover, we also have another fact that for any operator $A$ on
$\mathcal{X}$, we get

$A\xi=\sum_{i\in J}T_{i}^{\ast}\tilde{T}_{i}A\xi.$

That is, an association of operator frame and its dual frame can reconstruct
pointwisely an operator on $\mathcal{X}$ and so we can write

$A=\sum_{i\in J}T_{i}^{\ast}\tilde{T}_{i}A.$

For a frame of submodules $\\{W_{i}\\}_{i\in J}$ with respect to the family of
weights $\\{v_{i}\\}_{i\in J}$ for $\mathcal{X}$ with synthesis operator
$T_{W,v}$, the sequence $\\{u_{i}\\}_{i\in J}=\\{S_{W,v}^{-1}W_{i}\\}_{i\in
J}$ is called a dual frame of $\\{W_{i}\\}_{i\in J}$ with the operator
$S_{W,v}=T_{W,v}T_{W,v}^{\ast}$.

######  Theorem 6.4.

For a frame of submodules $\\{W_{i}\\}_{i\in J}$ with respect to the family of
weights $\\{v_{i}\\}_{i\in J}$ for $\mathcal{X}$, define
$T_{i}=v_{i}S_{W,v}\pi_{W_{i}}S_{W,v}^{-1}$ and
$Q_{i}=v_{i}\pi_{u_{i}}S_{W,v}^{-1}$ such that $S_{W,v}$ and its inverse are
uniformly bounded. Then $Q=\\{Q_{i}\\}_{i\in J}$ and $T=\\{T_{i}\\}_{i\in J}$
are all operator frames for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and $Q$ is
a dual frame of $T$.

###### Proof.

Assume that $\\{W_{i}\\}_{i\in J}$ has frame bounds $A$ and $B$.

  * (1)

$T=\\{T_{i}\\}_{i\in J}$ is an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.
For any $\xi\in\mathcal{X}$, we have

$\displaystyle\sum_{i\in J}\langle T_{i}\xi,T_{i}\xi\rangle$
$\displaystyle=\sum_{i\in J}\langle
v_{i}S_{W,v}\pi_{W_{i}}S_{W,v}^{-1}\xi,v_{i}S_{W,v}\pi_{W_{i}}S_{W,v}^{-1}\xi\rangle$
$\displaystyle\leq\|S_{W,v}\|_{\infty}^{2}B\langle
S_{W,v}^{-1}\xi,S_{W,v}^{-1}\xi\rangle$ $\displaystyle\leq
B\|S_{W,v}\|^{2}\|_{\infty}S_{W,v}^{-1}\|_{\infty}^{2}\langle\xi,\xi\rangle.$
On the other hand, $\displaystyle\sum_{i\in J}\langle
T_{i}\xi,T_{i}\xi\rangle$ $\displaystyle=\sum_{i\in J}\langle
v_{i}S_{W,v}\pi_{W_{i}}S_{W,v}^{-1}\xi,v_{i}S_{W,v}\pi_{W_{i}}S_{W,v}^{-1}\xi\rangle$
$\displaystyle\geq\|S_{W,v}^{-1}\|_{\infty}^{-2}A\langle
S_{W,v}^{-1}\xi,S_{W,v}^{-1}\xi\rangle$ $\displaystyle\geq
A\|S_{W,v}^{-1}\|_{\infty}^{-2}\|S_{W,v}\|_{\infty}^{-2}\langle\xi,\xi\rangle.$

Thus $T=\\{T_{i}\\}_{i\in J}$ is an operator frame.

  * (2)

$Q=\\{Q_{i}\\}_{i\in J}$ is also an operator frame for
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. The proof is similar to $(1)$.

  * (3)

If $T_{U,v}=S_{W,v}^{-1}T_{W,v}S_{W,v}$, then
$T_{U,v}^{\ast}=S_{W,v}^{-1}T_{W,v}^{\ast}S_{W,v}$ and $S_{U,v}=S_{W,v}$.
It is easy to check that $\pi_{u_{i}}=S_{W,v}^{-1}\pi_{W_{i}}S_{W,v}$.
Thus
$T_{U,v}=S_{W,v}^{-1}T_{W,v}S_{W,v},T_{U,v}^{\ast}=S_{W,v}^{-1}T_{W,v}^{\ast}S_{W,v}$
and so

$\displaystyle S_{U,v}$ $\displaystyle=T_{U,v}T_{U,v}^{\ast}$
$\displaystyle=S_{W,v}^{-1}T_{W,v}S_{W,v}S_{W,v}^{-1}T_{W,v}^{\ast}S_{W,v}$
$\displaystyle=S_{W,v}^{-1}T_{W,v}T_{W,v}^{\ast}S_{W,v}$
$\displaystyle=S_{W,v}^{-1}S_{W,v}S_{W,v}$ $\displaystyle=S_{W,v}.$ Hence, for
any $\xi\in\mathcal{X}$, we compute $\displaystyle\sum_{i\in
J}T_{i}^{\ast}Q_{i}\xi$ $\displaystyle=\sum_{i\in
J}v_{i}S_{W,v}^{-1}\pi_{W_{i}}S_{W,v}v_{i}S_{W,v}^{-1}\pi_{W_{i}}S_{W,v}S_{W,v}^{-1}x$
$\displaystyle=S_{W,v}^{-1}(\sum_{i\in J}v_{i}^{2}\pi_{W_{i}}\xi)$
$\displaystyle=S_{W,v}^{-1}S_{W,v}\xi$ $\displaystyle=\xi.$

This shows that $Q=\\{Q_{i}\\}_{i\in J}$ is an dual operator frame of the
operator frame $T=\\{T_{i}\\}_{i\in J}$.

This completes the proof. ∎

######  Theorem 6.5.

Assume that $\\{W_{i}\\}_{i\in J}$ is a Parseval frame of submodules for a
Hilbert $\mathcal{A}$-module $\mathcal{X}$. Then $\\{v_{i}\pi_{W_{i}}\\}_{i\in
J}$ is an operator frame for $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and a
dual frame of itself.

###### Proof.

When $\\{W_{i}\\}_{i\in J}$ is a Parseval frame of submodules, $S_{W,v}=I$.
Clearly, this theorem is a consequence of the last theorem. ∎

######  Theorem 6.6.

Let $\\{T_{i}\\}_{i\in I}$ and $\\{R_{j}\\}_{j\in J}$ be two operator frames
respectively in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and
$Hom_{\mathcal{A}}^{\ast}(\mathcal{Y})$, with duals $\\{\tilde{T_{i}}\\}_{i\in
I}$ and $\\{\tilde{R_{j}}\\}_{j\in J}$ respectively. Then
$\\{\tilde{T_{i}}\otimes\tilde{R_{j}}\\}_{i,j\in I,J}$ is a dual of
$\\{T_{i}\otimes R_{j}\\}_{i,j\in I,J}$.

###### Proof.

By definition, for all$x\in\mathcal{X}$ and $y\in\mathcal{Y}$, we have

$x=\sum_{i\in I}T_{i}^{\ast}\tilde{T_{i}}x,$ $y=\sum_{j\in
J}R_{j}^{\ast}\tilde{R_{j}}y$

and so

$(x\otimes y)=\sum_{i\in I}T_{i}^{\ast}\tilde{T_{i}}x\otimes\sum_{j\in
J}R_{j}^{\ast}\tilde{R_{j}}y,$ $\sum_{i\in
I}T_{i}^{\ast}\tilde{T_{i}}x\otimes\sum_{j\in
J}R_{j}^{\ast}\tilde{R_{j}}y=\sum_{i,j\in
I,J}T_{i}^{\ast}\tilde{T_{i}}x\otimes R_{j}^{\ast}\tilde{R_{j}}y,$ $\sum_{i\in
I}T_{i}^{\ast}\tilde{T_{i}}x\otimes\sum_{j\in
J}R_{j}^{\ast}\tilde{R_{j}}y=\sum_{i,j\in I,J}(T_{i}^{\ast}\otimes
R_{j}^{\ast}).(\tilde{T_{i}}x\otimes\tilde{R_{j}}y),$ $\sum_{i\in
I}T_{i}^{\ast}\tilde{T_{i}}x\otimes\sum_{j\in
J}R_{j}^{\ast}\tilde{R_{j}}y=\sum_{i,j\in I,J}(T_{i}\otimes
R_{j})^{\ast}.(\tilde{T_{i}}\otimes\tilde{R_{j}}).(x\otimes y).$

Thus $\\{\tilde{T_{i}}\otimes\tilde{R_{j}}\\}_{i,j\in I,J}$ is a dual of
$\\{T_{i}\otimes R_{j}\\}_{i,j\in I,J}$. ∎

######  Corollary 6.7.

Let $\\{T_{i,j}\\}_{0\leq i\leq n;j\in J}$ be a family of operator frames with
their duals $\\{\tilde{T}_{i,j}\\}_{0\leq i\leq n;j\in J}$ for $0\leq i\leq
n$. Then
$\\{\tilde{T}_{0,j}\otimes\tilde{T}_{1,j}\otimes......\otimes\tilde{T}_{n,j}\\}_{j\in
J}$ is a dual of $\\{T_{0,j}\otimes T_{1,j}\otimes......\otimes
T_{n,j}\\}_{j\in J}$.

## Declarations

Availablity of data and materials
Not applicable.

Competing interest
The authors declare that they have no competing interests.

Fundings
The authors declare that there is no funding available fot this paper.

Authors’ contributions
The authors equally conceived of the study, participated in its design and
coordination, drafted the manuscript, participated in the sequence alignment,
and read and approved the final manuscript.

## References

  * [1] L. Alizadeh and H. Mahmoud, On frames for countably generated Hilbert modules over locally $C^{\ast}$-algebras, Commun. Korean Math. Soc. 33 (2018), 527–533.
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11institutetext: Oxford Machine Learning in NeuroImaging (OMNI) Lab,
University of Oxford, UK 22institutetext: Wellcome centre for Integrative
Neuroimaging, FMRIB, University of Oxford, Oxford, UK 33institutetext:
Australian Institute for Machine Learning (AIML), Department of Computer
Science, University of Adelaide, Adelaide, Australia 44institutetext: South
Australian Health and Medical Research Institute (SAHMRI), North Terrace,
Adelaide, Australia
44email<EMAIL_ADDRESS>
# FedHarmony: Unlearning Scanner Bias with Distributed Data

Nicola K. Dinsdale 11 Mark Jenkinson 223344 Ana I.L. Namburete 11

###### Abstract

The ability to combine data across scanners and studies is vital for
neuroimaging, to increase both statistical power and the representation of
biological variability. However, combining datasets across sites leads to two
challenges: first, an increase in undesirable non-biological variance due to
scanner and acquisition differences - the harmonisation problem \- and second,
data privacy concerns due to the inherently personal nature of medical imaging
data, meaning that sharing them across sites may risk violation of privacy
laws. To overcome these restrictions, we propose FedHarmony: a harmonisation
framework operating in the federated learning paradigm. We show that to remove
the scanner-specific effects, we only need to share the mean and standard
deviation of the learned features, helping to protect individual subjects’
privacy. We demonstrate our approach across a range of realistic data
scenarios, using real multi-site data from the ABIDE dataset, thus showing the
potential utility of our method for MRI harmonisation across studies. Our code
is available at https://github.com/nkdinsdale/FedHarmony.

###### Keywords:

Harmonisation Federated Learning Domain Adaptation.

##  1 Introduction

Although some large scale projects, such as the UK Biobank, exist, the
majority of neuroimaging datasets remain small. This necessitates the
combination of data from multiple sites and scanners, both for statistical
power and to represent the breadth of biological variability. However,
combining data across scanners leads to an increase in non-biological
variance, due to differences such as acquisition protocol and hardware [9, 10,
24]. Thus, we need harmonisation methods to enable joint unbiased analysis of
data from different scanners and studies.

However, data cannot simply be pooled (combined) across imaging sites without
ensuring compliance with data privacy laws, particularly if we wish to use
representative clinical imaging data, as the sharing of medical images is
covered by legislation such as GDPR [2] and HIPAA [13]. Federated learning
(FL) has been proposed as a method to train models on distributed data [11],
where data are kept on their local servers, the users train local models on
their own private datasets and then share only the weights or gradients of the
trained models. FL has the potential to become the standard paradigm for
multisite imaging, with early studies demonstrating its feasibility in
neuroimaging [18].

Direct translation of existing harmonisation methods into FL frameworks is
non-trivial. Most deep learning methods for harmonisation are based on
generative frameworks [3, 12, 26, 27], and, although federated equivalents to
GANs and VAEs are being developed [16, 25], additional challenges exist for
harmonisation approaches that require simultaneous access to source and target
data [8]. Additionally, many methods require paired data – not possible with
distributed data and unlikely to exist in large multisite studies. Further,
most generative methods are data-hungry [3, 27], which casts into doubt
whether sufficient data would be available at local sites.

Alternative methods for harmonisation frame the task as a domain adaptation
problem (DA) [5], as the goal mirrors the harmonisation problem: removal of
information regarding domain (scanner or acquisition protocol) while retaining
the true variance of interest (the biological signal). In [5], the DA approach
has been demonstrated for a range of tasks and network architectures. The
harmonisation approach in [5] is built upon the DA framework proposed in [20]
and has since been translated into the federated setting [14]. In [14] the
domain adaptation across local sites is achieved by sharing feature embeddings
for local data points in a global shared knowledge bank, such that a global
domain predictor can be trained. However, sharing the features still leads to
privacy concerns [8], as images are potentially recoverable from the features.
Many of the other FL DA methods proposed cannot be adapted for the
harmonisation problem, as, for instance, they produce domain-specific models
or ensembles [8, 15, 21, 23], where the final predictions depend on the source
of the data.

Therefore, we present FedHarmony: a method to adapt the harmonisation
framework proposed in [5] for distributed data, while minimising shared
information. We aim to train a model that performs equally well on the task of
interest across all of the imaging sites while using features that are
invariant to the imaging site, through a horizontal federated framework [22].
We show that by approximating the learned feature embeddings as Gaussian
distributions, we can share only a mean and standard deviation per feature
which, by definition, contain no identifying information for individual
subjects. Alongside standard privacy-protecting approaches [7, 19], sharing
only these summary statistics would make the approach robust to honest-but-
curious adversaries [22]. By demonstrating our approach, FedHarmony, for a
range of data settings, we show its viability for data harmonisation for
multisite distributed imaging studies.

##  2 Methods

Figure 1: FedHarmony Procedure and high level general network architecture.

Consider the scenario in which we have multisite MRI data, where the data for
each site $h$ are stored on their local server, for all sites $H$. For each
site, we have pairs of training data $D_{h}=\\{\bm{X}_{h},\bm{y}_{h}\\}$ where
$\bm{X}_{h}$ is the input data and $\bm{y}_{h}$ is the label, and we consider
each site to be a separate data domain. We wish to create a model that
performs as well as possible on each imaging site, whilst having predictions
that are invariant to the acquisition site. A global model is constructed,
which following [5] is formed of a feature extractor (with weights
$\bm{\theta}_{repr}$), a label predictor ($\bm{\theta}_{p}$) and a domain
predictor ($\bm{\theta}_{d}$) (Fig. 1). This model is initialised randomly
(step 1), and then the FedHarmony procedure begins, as shown in Fig. 1, with
the global model weights for the three network components,
$\bm{\theta}_{repr,t}$, $\bm{\theta}_{p,t}$ and $\bm{\theta}_{d,t}$, being
sent to the local sites; $t$ indicates these are the current weights for
iteration $t$ and thus each site receives the same initialisation (step 2).

Update Global Knowledge Store (Step 3): To remove scanner information
adversarially, the different sites’ nodes need to have an understanding of the
various scanner characteristics. Although there is precedent for achieving
this through sharing the feature embeddings from local sites to the global
site [14], we aim to minimise the quantity of shared information to help
protect individual privacy. It has been shown that domain-specific
characteristics can be encoded to enable contrastive learning [6], by sharing
the mean and standard deviation of the feature embeddings following a BoxCox
transformation [1] (an invertible transform which aims to make the
distribution maximally Gaussian), suggesting that this might be sufficient
information to also remove scanner information. However, in many multisite MRI
studies, small amounts of local data are available that may be insufficient to
fit a BoxCox transformation. Thus, we simplify the transformation by assuming
that the features will already be normally distributed and that we can
characterise $\bm{Q}_{h}=featureExtractor(\bm{X}_{h},\bm{\theta}_{repr,t})$ by
their mean and standard deviations. Therefore the total information shared per
site is a mean and standard deviation per feature, which inherently has no
information about individuals, and we collate a global information store
$\\{\bm{\mu}_{h},\bm{\sigma}_{h}:h\in H\\}$.

Training Procedure (Step 4): Local training is then controlled by three loss
functions, based on [5]. First is the main task loss $L_{p}$, in which the
feature extractor and the label predictor are updated. If
$\Theta_{t}=[\bm{\theta}_{repr,t},\bm{\theta}_{p,t}]$ are the current global
weights (iteration $t$) for the feature extractor and the label predictor, and
$\Theta_{t+1}^{h}=[\bm{\theta}_{repr,t+1}^{h},\bm{\theta}_{p,t+1}^{h}]$ are
the updated ($t+1$) weights for site $h$, then:

$L_{p}(\bm{X}_{h},\bm{y}_{h};\bm{\theta}_{repr,t+1}^{h},\bm{\theta}_{p,t+1}^{h})=\frac{1}{N_{h}}\sum_{j=1}^{N_{h}}L_{task}(\bm{y}_{h,j},\hat{\bm{y}}_{h,j})+\mu
L_{prox}(\Theta_{t},\Theta_{t+1}^{h})$ (1)

where $L_{task}$ is the task-specific loss function, averaged over the
training samples available for the given site, $N_{h}$, and $\bm{y}_{h,j}$ and
$\hat{\bm{y}}_{h,j}$ are the true and predicted labels for the $j^{th}$
example for the $h^{th}$ imaging site. $L_{prox}$ is a proximal loss term
based on FedProx [17] that penalises weight deviations from the global model
that do not sufficiently improve the main task loss and has been shown to aid
FL when the data is non-iid [17], as is inherently true for multisite MRI
data, hence the harmonisation problem. Thus, if $\mu$ is a constant, weighting
the contributions from the two losses, then:

$L_{prox}(\Theta_{t},\Theta_{t+1}^{h})=||\bm{\theta}_{repr,t}-\bm{\theta}_{repr,t+1}^{h}||^{2}+||\bm{\theta}_{p,t}-\bm{\theta}_{p,t+1}^{h}||^{2}$
(2)

The second loss function, $L_{d}$, then updates the domain predictor to be
able to discriminate between the sites. Data is only available for the local
site $h$ and thus can only directly generate the features, $\bm{Q}_{h}$,
needed to train the domain predictor for this local site. For all the other
sites $\breve{H}$ we generate example features randomly using the shared
knowledge store; thus, for site $\breve{h}$ in
$\breve{H}:\bm{\widetilde{Q}}_{\breve{h}}\sim\mathcal{N}(\bm{\mu}_{\breve{h}},\bm{\sigma}_{\breve{h}})$.
The generated features,
$\bm{\widetilde{Q}}_{\breve{H}}=\\{\bm{\widetilde{Q}}_{\breve{h}}\forall\breve{h}\in\breve{H}\\}$
are then concatenated with the true features, $\bm{Q}_{h}$, such that
$\bm{\widetilde{Q}}_{H}^{h}=[\bm{Q}_{h},\bm{\widetilde{Q}}_{\breve{H}}]$. The
number of simulated subjects generated for each site $\breve{h}$ is chosen
randomly such that the final batchsize is twice that of the original; the
features and corresponding domain (site) labels are then shuffled. We then
update the parameters of the domain predictor, using categorial cross-entropy
to assess the site information remaining in $\bm{Q}_{h}$:

$L_{d}(\bm{\widetilde{Q}}^{h}_{H},\bm{d};\bm{\theta}_{d,t+1}^{h})=-\sum^{H}_{h=1}\mathbbm{1}[d=h]log(p_{h})$
(3)

where $p_{h}$ are the softmax outputs of the domain classifier. The third loss
is the confusion loss $L_{conf}$, which removes the site information, by
penalising deviation in $p_{h}$ from a uniform distribution:

$L_{conf}(\bm{X}_{h},\bm{\theta}_{d,t+1}^{h};\bm{\theta}_{repr,t+1}^{h})=-\sum^{H}_{h=1}\frac{1}{H}log(p_{h})$
(4)

Therefore, the overall method can be considered to minimise the total loss
function $L=L_{p}+\alpha L_{d}+\beta L_{conf}$ where $\alpha$ and $\beta$
control the relative contributions of the different loss functions. The domain
loss (Eq. 3) and the confusion loss (Eq. 4) act in opposition to each other,
and so must be updated iteratively, with three iterations per epoch. The
procedure is iterated through for $E$ local epochs.

Weight Aggregation (Step 5): After each site has completed the $E$ local
epochs, the local weights are returned to the global node to be aggregated. In
[5] it was shown that the harmonisation process was aided through evaluating
each site separately in the loss function rather than averaging over all data
points. Given that FedAvg [11] is equivalent to averaging over all training
subjects for FL, we alter the aggregation to have equal contributions from
each site:
$\bm{\theta}_{t+1}\leftarrow\frac{1}{H}\sum^{H}_{h=1}\bm{\theta}_{t+1}^{h}$
where the aggregation is completed separately for $\bm{\theta}_{repr}$,
$\bm{\theta}_{p}$ and $\bm{\theta}_{d}$. We term this aggregation step
FedEqual. Once the aggregation is complete, the training loops through stages
2 - 5 (Fig. 1) until convergence.

Semi-supervised setting: We then explore the scenario where labels are only
available for a subset of sites, which is realistic, as labels are expensive
to generate in terms of time and expertise. Thus, we also consider the semi-
supervised setting, adjusting the framework such that local training is only
completed for sites with labels available, but we continue to update the
global knowledge store for all sites. Therefore, the removal of the site
information still considers all sites.

##  3 Implementation Details

For our experiments we use data from the ABIDE dataset111Data from:
https://fcon_1000.projects.nitrc.org/indi/abide/ [4], for the task of age
prediction, using T1 MR images as $\bm{X}$ and the ages as the labels y. The
MR images from each site were processed using FSL
anat222https://fsl.fmrib.ox.ac.uk/fsl/fslwiki/fsl_anat. Subjects were rejected
where the pipeline failed or where data was missing. Four sites (Trinity, NYU,
UCLA, Yale) were chosen for our experiments, so as to span both age
distributions and subject numbers (Fig. 2). The data were split into
training/validation/test sets as $70\%/10\%/20\%$, yielding a maximum of 127
subjects for training (NYU) and a minimum of 35 (Trinity). Our networks and
baselines were all trained with 5-fold cross-validation until no improvement
in the average mean absolute error (MAE) was reported on the validation data
for all local sites. We present the results on the held-out testing data in
all cases.

Figure 2: Normalised age distributions for the 4 sites from the ABIDE dataset:
Trinity: 49 subjects, 16.7$\pm$3.6 years (range: 12-25), NYU: 182 subjects,
14.7$\pm$6.6 (6-39), UCLA: 99 subjects, 12.5$\pm$2.2 (8-17), Yale: 56
subjects, 12.2$\pm$2.8 (7-17).

For the network, we used the VGG-based architecture used in [5]333code from:
https:/github.com/nkdinsdale/Unlearning_for_MRI_harmonisation. The images were
resized to (128, 240, 160) and normalised to have zero mean and unit standard
deviation. As demonstrated by [5], the framework is flexible and should be
applicable to many feedforward architectures and tasks, with age prediction
being used to demonstrate the approach due to the availability of labels
across many sites. The implementation was in Python 3.6.8 and PyTorch 1.10.2.
The networks were trained on a 16GB P100 GPU, taking up to 5 minutes per local
epoch (NYU), with a batchsize of 16, and requiring 50-75 rounds of
communication to converge (FedHarmomy required 61 rounds and FedAvg required
56). The following hyperparameters were used for all experiments based on [5,
17], with the same values applied for each local site: batchsize=16, learning
rate=$1\times 10^{-4}$ with Adam optimiser, local epochs (E)=10, $\mu$=0.01,
$\alpha$=1, and $\beta$=100.

##  4 Results & Discussion

Fully Supervised: We first compared our method to several baselines: training
on NYU data only (the largest site, representing where data cannot be shared
and we have no federated learning framework), training on all sites normally
(centralised data), FedAvg [11], FedProx [17] and our proposed FedEqual
aggregation method. We also completed an ablation study, exploring the effect
of the aggregation and proximal losses alongside the harmonisation. To compare
methods, we evaluated both the mean absolute error (MAE), where we want the
best performance possible across all sites, and the scanner classification
accuracy (SCA): the performance of a domain classifier trained on $\bm{Q}$ at
convergence, where the aim is to achieve random chance (25%). Note that the
domain predictor was able to achieve 95% on the feature embeddings when the
age prediction was trained with NYU only, and so is clearly able to identify
scanner information.

Considering the results in Table 4, we were able train federated models
(FedAvg) to perform as well as standard centralised training. FedHarmony led
to improvement both in terms of average performance across all sites and in
the reduction of SCA to close to random chance. This improvement indicates the
success of the DA approach, as information from across sites is being used to
aid predictions even though only summary statistics of the features are being
shared. The ablation study shows the need for the three additions to the
FedAvg baseline, with the $L_{prox}$ (Eq. 2) being vital for both stability of
training and the ability to remove scanner information. Fig. 3 shows a PCA of
$\bm{Q}$ comparing FedAvg and FedHarmony, where it is evident that the
harmonisation increases the feature overlap of the different sites.

Table 1: Fully supervised results: Component represents additions relative to
FedAvg [11]: Equal = Aggregation as FedEqual; Prox = Proximal Loss as Eq. 2;
Harm = Harmonisation, otherwise local training only considers the main task
loss. SCA = Scanner Classification Accuracy of domain classifier retrained on
$\bm{Q}$ at the end of training, where random chance (25%) is the goal. NYU
only and all sites = centralised training. FedHarmony is our proposed
approach. * = significant improvement over next performing method (paired
t-test, $p<0.001$).

Method | Component | Site MAE |  Average MAE |  SCA (%)
---|---|---|---|---
Equal | Prox | Harm | NYU n=40 | Yale n=14 | UCLA n=20 | Trinity n=10
NYU Only |  |  |  | 5.26$\pm$4.37 | 2.38$\pm$1.42 | 3.22$\pm$2.26 | 9.66$\pm$7.89 | 5.13 | 95
All Sites |  |  |  | 5.10$\pm$4.56 | 2.32$\pm$1.50 | 2.87$\pm$2.13 | 4.53$\pm$3.49 | 3.70 | 86
FedAvg [11] |  |  |  | 5.26$\pm$4.74 | $\bm{1.99\pm 1.21}^{*}$ | 2.57$\pm$1.88 | 4.94$\pm$3.80 | 3.69 | 64
FedProx [17] |  | ✓ |  | 5.21$\pm$4.70 | 2.15$\pm$1.44 | 2.61$\pm$1.73 | 4.47$\pm$3.56 | 3.61 | 62
FedEqual | ✓ |  |  | 6.07$\pm$3.99 | 2.06$\pm$1.31 | 2.61$\pm$2.30 | 5.43$\pm$3.00 | 4.06 | 42
Ablation A |  |  | ✓ | 12.21$\pm$10.31 | 4.20$\pm$3.61 | 6.21$\pm$3.21 | 10.32$\pm$8.31 | 8.23 | 95
Ablation B | ✓ |  | ✓ | 5.27$\pm$4.90 | 2.27$\pm$1.13 | 2.66$\pm$1.98 | 4.81$\pm$3.98 | 3.75 | 34
Ablation C |  | ✓ | ✓ |  $4.93\pm 4.31$^* | 2.13$\pm$1.91 | 2.26$\pm$1.15 | 4.36$\pm$3.37 | $\bm{3.42}$ | 35
FedHarmony | ✓ | ✓ | ✓ | 5.14$\pm$4.52 | 2.04$\pm$1.25 | $\bm{2.13\pm 1.45}^{*}$ | $\bm{4.34\pm 3.65}$ | $\bm{3.42}$ | $\bm{29}$

(a) FedAvg

(b) FedHarmony

Figure 3: PCA of $\bm{Q}$ for FedAvg and FedHarmony, showing, as expected,
increased overlap across sites with FedHarmony.

Table 2: Semi-supervised results: NYU only = centralised training only NYU; No
Trinity = centralised training no Trinity; All sites = centralised training.
SCA = Scanner Classification Accuracy of domain classifier retrained on
$\bm{Q}$ at the end of training, where random chance (25%). FedHarmony was
trained locally for only labelled sites, and global shared information used
for all sites. * = significant improvement over next performing method (paired
t-test, $p<0.001$).

Method | Site MAE |  Average MAE |  SCA (%)
---|---|---|---
NYU n=40 | Yale n=14 | UCLA n=20 | Trinity n=10
a) Semi-supervised 1 Site
NYU Only | $\bm{5.26\pm 4.37}$ | 3.22$\pm$2.26 | 3.22$\pm$2.26 | 9.66$\pm$7.89 | 5.13 | 95
FedHarmony | $\bm{5.26\pm 4.15}$ | $\bm{2.20\pm 1.79}^{*}$ | $\bm{2.91\pm 1.56}^{*}$ | $\bm{4.09\pm 3.32^{*}}$ | 3.61∗ | 30
b) Semi-supervised 3 Site
No Trinity | 5.36$\pm$4.65 | $\bm{1.81\pm 1.66}^{*}$ | $\bm{1.83\pm 1.16}^{*}$ | 7.29$\pm$5.00 | 4.07 | 96
FedProx [17] | 5.15$\pm$4.76 | 2.15$\pm$1.38 | 2.62$\pm$1.82 | 5.96$\pm$4.05 | 3.96 | 74
FedHarmony | $\bm{5.12\pm 4.60}^{*}$ | 2.39$\pm$1.67 | 2.16$\pm$1.40 | $\bm{4.69\pm 4.19}^{*}$ | 3.59∗ | 26
All Sites | 5.10$\pm$4.56 | 2.32$\pm$1.50 | 2.87$\pm$2.13 | 4.53$\pm$3.49 | 3.70 | 86

Semi-supervised: We then simulated two semi-supervised scenarios, where
training labels were only available for a subset of the sites. First we
considered when only one site had available training labels, NYU, as a
multitarget DA task, which could for example represent when the organising
site has labelled data. In Table 2 we compare to standard training (FedAvg and
FedProx are equivalent to standard training for a single labelled site). Using
FedHarmony, we were able to train the network to perform as well as fully
supervised centralised training, while removing the scanner information,
showing the power of the shared global knowledge. We also considered when
three sites have labels and one does not (Trinity), for example, if a new site
joins the study. FedHarmony led to improvement in performance again, even
compared to having labels for all sites, while removing the scanner
information. These results show the suitability of our approach across
realistic data scenarios for multisite MRI studies.

Suitability of Gaussian Fit: In Fig. 4 we plot the standard deviation of 100
mean estimates for $\bm{Q}_{NYU}$ from both the direct Gaussian fit and the
BoxCox transformation for increasing numbers of samples. It is evident that
the direct Gaussian fit leads to more consistent estimates, indicating
increased suitability in low data regimes compared to the BoxCox transform, as
the representation on the training data is more likely to also represent the
testing data.

Figure 4: We plot the standard deviation of 100 mean estimates of the fit for
$\bm{Q}_{NYU}$, for increasing numbers of random samples from the full
$\bm{Q}_{NYU}$, alongside consistent and inconsistent fits. Note: the
estimated means will differ, as the BoxCox transforms the data to be maximally
Gaussian rather than fitting the data.

##  5 Conclusion

We have presented FedHarmony, a method to allow harmonisation of MRI data in a
federated learning scenario, through development of a domain adaptation
approach with minimal information sharing, outperforming baseline and FL
approaches across realistic data scenarios. Alongside standard privacy-
protecting approaches, such as differential privacy, our approach would enable
training models on multisite MRI data while maintaining individual subjects’
privacy.

##  6 Acknowledgements

ND is supported by a Academy of Medical Sciences Springboard Award. MJ is
supported by the National Institute for Health Research, Oxford Biomedical
Research Centre, and this research was funded by the Wellcome Trust
[215573/Z/19/Z]. WIN is supported by core funding from the Wellcome Trust
[203139/Z/16/Z]. AN is grateful for support from the UK Royal Academy of
Engineering under the Engineering for Development Research Fellowships scheme
and the Academy of Medical Sciences. The computational aspects of this
research were supported by the Wellcome Trust Core Award [Grant Number
203141/Z/16/Z] and the NIHR Oxford BRC. The views expressed are those of the
author(s) and not necessarily those of the NHS, the NIHR or the Department of
Health.

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