# Strong Gelfand Pairs of SL($\bf{2,p}$)

Andrea Barton, Stephen P. Humphries  Department of Mathematics, Brigham Young
University, Provo, UT 84602, U.S.A. E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

A strong Gelfand pair $(G,H)$ is a group $G$ together with a subgroup $H$ such
that every irreducible character of $H$ induces a multiplicity-free character
of $G$. We classify the strong Gelfand pairs of the special linear groups
$SL(2,p)$ where $p$ is a prime.

Keywords: Strong Gelfand pair, special linear group, upper triangular group,
characters, finite groups.

###### 2010 Mathematics Subject Classification:

Primary: 20G40 Secondary: 20C15 20G05

##  1\. Introduction

Given a finite group $G$, a multiplicity-free character of $G$ is a character
$\chi$ of $G$ such that for every irreducible character $\psi$ of $G$, we have
$\langle\chi,\psi\rangle\leq 1$.

A Gelfand pair $(G,H)$ is a group $G$ together with a subgroup $H$ such that
the trivial character of $H$ induces a multiplicity-free character of $G$. The
importance of Gelfand pairs is indicated by the following equivalent
conditions for finite groups ([3], [19]):

(i) The algebra of $(H,H)$-double invariant functions on G with multiplication
defined by convolution is commutative.

(ii) For any irreducible representation $\rho$ of $G$, the space of
$H$-invariant vectors in $\rho$ is no more than 1-dimensional.

(iii) For any irreducible representation $\rho$ of $G$, the dimension of ${\rm
Hom}_{H}(\rho,\mathbf{1})$ is less than or equal to 1, where $\mathbf{1}$
denotes the trivial representation.

(iv) The permutation representation of $G$ on the cosets of $H$ is
multiplicity-free.

(v) The centralizer algebra of the permutation representation is commutative.

(vi) The double cosets $HgH$ determine a commutative Schur ring over $G$.

A strong Gelfand pair $(G,H)$ is a finite group $G$ together with a subgroup
$H$ such that every irreducible character of $H$ induces a multiplicity-free
character of $G$. In other words, for any $\chi\in\hat{G}$, $\psi\in\hat{H}$,
where $\hat{G}$ is the set of all irreducible characters of $G$ and $\hat{H}$
is the set of all irreducible characters of $H$, we have
$\langle\chi,\psi\uparrow^{G}\rangle\leq 1$. We note that a strong Gelfand
pair is also referred to using the phrase “multiplicity one property” or
“multiplicity one theorem” (see [1, 2, 16]).

Equivalently, $(G,H)$ is a strong Gelfand pair if and only if the Schur ring
determined by the $H$-classes $g^{H}=\\{g^{h}:h\in H\\},g\in G$, is
commutative [5].

The strong Gelfand subgroups of the symmetric group were found in [3]; those
for wreath products of the form $F\wr S_{n}$, where $F$ is a finite abelian
group, were found in [4].

Let $G={\rm SL}(2,p)$ and let
$U=U(2,p)=\mathcal{C}_{p-1}\rtimes\mathcal{C}_{p}\leq G$ be the subgroup of
upper triangular matrices. When $p$ is odd, we let $H_{2}\leq U$ be the unique
subgroup of $U$ of index $2$. The main result of this paper is to determine
the strong Gelfand pairs of ${\rm SL}(2,p)$:

######  Theorem 1.1.

Let $p>11$ be prime.

(i) If $p\equiv 1\operatorname{mod}4$, then there is only one strong Gelfand
pair $(G,U)$.

(ii) If $p\equiv 3\operatorname{mod}4$, then there are two strong Gelfand
pairs: $(G,U)$ and $(G,H_{2})$.

In Theorem 4.4 we classify the strong Gelfand pairs for primes $p\leq 11$.

Acknowledgment All computations were made using Magma [18].

##  2\. Character Tables

In this section we give the character tables for $G$, $PSL(2,p)$, and $U$.
This involves finding the character table for certain semidirect products. Let
$p$ be a prime.

From Suzuki [17, p. 393], the conjugacy classes of $U$ have the following
representatives:

(i) $1=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$;

(ii) $-1=\begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix}$;

(iii) Let $a=\begin{pmatrix}\lambda&0\\\ 0&1/\lambda\end{pmatrix}$, where
$\lambda$ is a generator of $\mathbb{F}_{p}^{*}$. Then each $a^{i}$ represents
a distinct class for $1\leq i<p-1$, $i\neq(p-1)/2$;

(iv) $b=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$;

(v) $d=\begin{pmatrix}1&t\\\ 0&1\end{pmatrix}$, where $t$ is a non-square mod
$p$;

(vi) $-b=\begin{pmatrix}-1&-1\\\ 0&-1\end{pmatrix}$;

(vii) $-d=\begin{pmatrix}-1&-t\\\ 0&-1\end{pmatrix}$.

Thus there are $p+3$ classes. The sizes of these classes are:

(i) 1; (ii) 1; (iii) $p$; (iv) $(p-1)/2$;

(v) $(p-1)/2$; (vi) $(p-1)/2$; (vii) $(p-1)/2$.

Observe that $U=\langle a,b\rangle$. We will use this to develop the character
table.

Let $B=\langle b\rangle$, the subgroup of $U$ whose diagonal entries are 1.
Then $B\triangleleft U$, so we lift the characters of $U/B\cong\langle
a\rangle$ to get $p-1$ linear characters $1_{U},\chi_{0,k}$ for $1\leq k\leq
p-2$ which are trivial on $B$. Let $\zeta$ be a primitive $(p-1)^{\text{th}}$
root of unity. These linear characters are determined by their values on the
generators $a$ and $b$:

$\chi_{0,k}(a)=\zeta^{k},\quad\chi_{0,k}(b)=1.$

Because $U$ is a split metacyclic group, we can use a result of Munkholm [13]
to complete the character table as shown below.

Let $B^{\prime}=\langle-1,b\rangle$. To find the nonlinear characters of $U$,
we induce the linear characters of $B^{\prime}$. These linear characters are
given by

$T_{i,k}(b)=\eta^{i},\quad T_{i,k}(-1)=(-1)^{k},$

where $\eta$ is a primitive $p^{\text{th}}$ root of unity, $1\leq i<p$, and
$k=1,2$. Recall that $\lambda$ is the generator of $\mathbb{F}_{p}^{*}$ used
in the definition of $a$. The characters are induced as follows, from [13, p.
458]:

$\displaystyle\chi_{i,k}((-1)^{\nu}b^{\tau})$
$\displaystyle=\sum_{\ell=0}^{(p-3)/2}T_{i,k}(a^{\ell}((-1)^{\nu}b^{\tau})a^{-\ell})=\sum_{\ell=0}^{(p-3)/2}T_{i,k}((-1)^{\nu}b^{\tau\lambda^{2\ell}})$
$\displaystyle=\sum_{\ell=0}^{(p-3)/2}(-1)^{\nu
k}\eta^{i\tau\lambda^{2\ell}}=(-1)^{\nu
k}\sum_{\ell=0}^{(p-3)/2}\eta^{i\tau\lambda^{2\ell}}.$

Let

$Z=\chi_{1,1}(b)=\Sigma_{\ell=0}^{(p-3)/2}\eta^{\lambda^{2\ell}}$

and

$Z^{(t)}=\chi_{1,1}(b^{t})=\Sigma_{\ell=0}^{(p-3)/2}\eta^{t\lambda^{2\ell}},$

where $t$ is the non-square mod $p$ used in the definition of $d$.

When $k$ is a square $\operatorname{mod}p$, each $T_{i,k}$ induces the
character $\chi_{i,1}$. Additionally, when $k$ is a non-square
$\operatorname{mod}p$, each $T_{i,k}$ induces the character $\chi_{i,2}$. By
the Quadratic Gauss sum formula [6, p. xxvii]:

######  Lemma 2.1.

If $p\equiv 1\operatorname{mod}4$, then $Z=\frac{1}{2}(-1+\sqrt{p}),$ and
$Z^{(t)}=\frac{1}{2}(-1-\sqrt{p})$.

If $p\equiv 3\operatorname{mod}4$, then $Z=\frac{1}{2}(-1+i\sqrt{p})$ and
$Z^{(t)}=\frac{1}{2}(-1-i\sqrt{p})$.

Then we have

$\chi_{1,1}(-b)=-Z\quad\text{and}\quad\chi_{1,1}(-b^{t})=-{Z}^{(t)}.$

One then sees that

$\chi_{1,2}(b)=Z,\quad\chi_{1,2}(-b)=Z,\quad\chi_{1,2}(b^{t})={Z}^{(t)},\quad\text{and}\quad\chi_{1,2}(-b^{t})={Z}^{(t)}.$

This gives the following character table for $U$:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b$ | $-d$
---|---|---|---|---|---|---|---
$1$ | 1 | 1 | 1 | 1 | 1 | 1 | 1
$\chi_{0,k}$ | $1$ | $(-1)^{k}$ | $\zeta^{kj}$ | 1 | 1 | $(-1)^{k}$ | $(-1)^{k}$
$\chi_{1,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$
$\chi_{1,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$
$\chi_{2,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$
$\chi_{2,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$
Table 1. Character Table of $U$

We continue using notation from above, and now let $\sigma$ be a primitive
$(p+1)^{\text{th}}$ root of 1. Then a result of Frobenius [9, p. 30] tells us
that the (ordinary) complex character table of $G$, with $p\equiv
1\operatorname{mod}4$, is as shown in Table 2 for $1\leq i\leq(p-3)/2$, $1\leq
j\leq(p-1)/2$, $1\leq l\leq(p-3)/2$, and $1\leq m\leq(p-1)/2$.

| 1 | -1 | $b$ | $d$ | $-b$ | $-d$ | $a^{l}$ | $f^{m}$
---|---|---|---|---|---|---|---|---
$1$ | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1
$\varphi$ | $p$ | $p$ | 0 | 0 | 0 | 0 | 1 | $-1$
$\chi_{i}$ | $p+1$ | $(-1)^{i}(p+1)$ | 1 | 1 | $(-1)^{i}$ | $(-1)^{i}$ | $\zeta^{il}+\zeta^{-il}$ | 0
$\theta_{j}$ | $p-1$ | $(-1)^{j}(p-1)$ | $-1$ | $-1$ | $(-1)^{j+1}$ | $(-1)^{j+1}$ | $0$ | $-(\sigma^{jm}+\sigma^{-jm})$
$\xi_{1}$ | $(p+1)/2$ | $(p+1)/2$ | $-Z^{(t)}$ | $-Z$ | $-Z^{(t)}$ | $-Z$ | $(-1)^{l}$ | 0
$\xi_{2}$ | $(p+1)/2$ | $(p+1)/2$ | $-Z$ | $-Z^{(t)}$ | $-Z$ | $-Z^{(t)}$ | $(-1)^{l}$ | 0
$\eta_{1}$ | $(p-1)/2$ | $-(p-1)/2$ | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$ | 0 | $(-1)^{m+1}$
$\eta_{2}$ | $(p-1)/2$ | $-(p-1)/2$ | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$ | 0 | $(-1)^{m+1}$
Table 2. Character Table of $G$, with $p\equiv 1\operatorname{mod}4$

In the case $p\equiv 3\operatorname{mod}4$, the final four characters are
different, as shown in Table 3, where $Z$ and $Z^{(t)}$ for $p\equiv
3\operatorname{mod}4$ are as given above.

| 1 | -1 | $b$ | $d$ | $-b$ | $-d$ | $a^{l}$ | $f^{m}$
---|---|---|---|---|---|---|---|---
$\xi_{1}^{\prime}$ | $(p+1)/2$ | $-(p+1)/2$ | $-Z^{(t)}$ | $-Z$ | $Z^{(t)}$ | $Z$ | $(-1)^{l}$ | 0
$\xi_{2}^{\prime}$ | $(p+1)/2$ | $-(p+1)/2$ | $-Z$ | $-Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $(-1)^{l}$ | 0
$\eta_{1}^{\prime}$ | $(p-1)/2$ | $(p-1)/2$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | 0 | $(-1)^{m+l}$
$\eta_{2}^{\prime}$ | $(p-1)/2$ | $(p-1)/2$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$ | 0 | $(-1)^{m+l}$
Table 3. Characters of $G$, with $p\equiv 3\operatorname{mod}4$

From the character table of ${\rm SL}(2,p)$ (Tables 2, 3) we deduce the
following character table for ${\rm PSL}(2,p)$, which can also be found in
[7]; see Table 4, where $1\leq i\leq(p-3)/2$, $1\leq j\leq(p-1)/2$, $1\leq
l\leq(p-3)/2$, and $1\leq m\leq(p-1)/2$. Here, we also require $i$ and $j$ to
be even.

| 1 | $b$ | $d$ | $a^{l}$ | $f^{m}$
---|---|---|---|---|---
$1$ | 1 | 1 | 1 | 1 | 1
$\overline{\varphi}$ | $p$ | 0 | 0 | 1 | $-1$
$\overline{\chi_{i}}$ | $p+1$ | 1 | 1 | $\zeta^{il}+\zeta^{-il}$ | 0
$\overline{\theta_{j}}$ | $p-1$ | $-1$ | $-1$ | $0$ | $-(\sigma^{jm}+\sigma^{-jm})$
$\overline{\xi_{1}}$ | $(p+1)/2$ | $-Z^{(t)}$ | $-Z$ | $(-1)^{l}$ | 0
$\overline{\xi_{2}}$ | $(p+1)/2$ | $-Z$ | $-Z^{(t)}$ | $(-1)^{l}$ | 0
Table 4. Character table for ${\rm PSL}(2,p)$ with $p\equiv
1\operatorname{mod}4$

In the case $p\equiv 3\operatorname{mod}4$, the last two characters
$\overline{\xi_{1}},\overline{\xi_{2}}$ are different, as indicated in Table
5.

| 1 | $b$ | $d$ | $a^{l}$ | $f^{m}$
---|---|---|---|---|---
$\overline{\eta_{1}^{\prime}}$ | $(p-1)/2$ | $Z$ | $Z^{(t)}$ | 0 | $(-1)^{m+l}$
$\overline{\eta_{2}^{\prime}}$ | $(p-1)/2$ | $Z^{(t)}$ | $Z$ | 0 | $(-1)^{m+l}$
Table 5. Characters of ${\rm PSL}(2,p)$, with $p\equiv 3\operatorname{mod}4$

##  3\. Group Theory Lemmas

We collect here results that will be used in the proof of Theorem 1.1.

######  Lemma 3.1.

Suppose $H\leq K\leq G$ are groups. If $(G,K)$ is not a strong Gelfand pair,
then neither is $(G,H)$.

###### Proof.

Suppose $(G,K)$ is not a strong Gelfand pair. So there are
$\chi\in\hat{G},\psi\in\hat{K}$ with $\langle\chi,\psi\uparrow^{G}\rangle\geq
2$. Let $\varphi$ be an irreducible character of H with
$\langle\varphi\uparrow^{K},\psi\rangle>0.$ Because
$(\varphi\uparrow^{K})\uparrow^{G}=\varphi\uparrow^{G}$ and by Frobenius
reciprocity, we have:

$\displaystyle\langle\varphi\uparrow^{G},\chi\rangle$
$\displaystyle=\langle(\varphi\uparrow^{K})\uparrow^{G},\chi\rangle=\langle\varphi\uparrow^{K},\chi|_{K}\rangle$
$\displaystyle\geq\langle\varphi\uparrow^{K},\psi\rangle\cdot\langle\psi,\chi|_{K}\rangle>1.$

Thus, $(G,H)$ is not a strong Gelfand pair. ∎

######  Lemma 3.2.

Suppose $N\leq H\leq G$ are groups with $N\triangleleft G$. Then $(G,H)$ is a
strong Gelfand pair if and only if $(G/N,H/N)$ is a strong Gelfand pair.

###### Proof.

This result follows from the fact that the lift of an irreducible character is
irreducible [12, Theorem 17.3]. ∎

For a finite group $G$, the total character $\tau$ of $G$ is the sum of all
the irreducible characters of $G$ [10, 14, 15].

######  Lemma 3.3.

Let $H\leq G$ and let $\tau$ be the total character of $H$. If there is some
character $\chi$ of $G$ with $\deg(\tau)<\deg(\chi)$, then $(G,H)$ is not a
strong Gelfand pair.

###### Proof.

Let $\varphi_{1},\cdots,\varphi_{r}$ be the irreducible characters of $H$.
Then for any character $\chi$ of $G$ we have

$\chi|_{H}=\Sigma_{i=1}^{r}\epsilon_{i}\varphi_{i},$

where each $\epsilon_{i}\in\mathbb{N}$. Assume $(G,H)$ is a strong Gelfand
pair. So, $\epsilon_{i}\leq 1$ for all $i$. Thus

$\deg(\chi)=\Sigma_{i=1}^{r}\epsilon_{i}\deg(\varphi_{i})\leq\Sigma_{i=1}^{r}\deg(\varphi_{i})=\deg(\tau).$

We conclude that whenever $\deg(\tau)<\deg(\chi)$, it must be that $(G,H)$ is
not a strong Gelfand pair. ∎

######  Lemma 3.4.

There is a one-to-one correspondence between maximal subgroups of ${\rm
SL}(2,p)$ and maximal subgroups of ${\rm PSL}(2,p)$.

###### Proof.

By the lattice isomorphism theorem, this is equivalent to showing each maximal
subgroup of $G$ contains $Z(G)=\langle-1\rangle$. By way of contradiction,
suppose there is a maximal subgroup $M$ of $G$ such that $-1\notin M$. Then
$a\notin M$, because $a^{(p-1)/2}=-1$. But then, $a\notin\langle-1,M\rangle$,
so $M$ was not maximal. ∎

##  4\. Proof of Theorem 1.1

We first slightly restate Theorem 1.1:

######  Theorem 4.1.

Let $G={\rm SL}(2,p)$ and let $U\leq G$ denote the upper triangular subgroup.
Let $H_{2}\leq U$ denote the subgroup of elements whose diagonal entries are
(non-zero) squares in $\mathbb{F}_{p}$.

For prime $p>11$, if $p\equiv 1\operatorname{mod}4$, then there is only one
strong Gelfand pair $(G,U)$, while if $p\equiv 3\operatorname{mod}4$, then
there are two, $(G,U)$ and $(G,H_{2})$.

The cases for $p<13$ are given in Theorem 4.4.

In this section, we now assume $p\geq 13$.

By Lemma 3.1, we only need to consider maximal subgroups of ${\rm SL}(2,p)$.
To do this, we use Lemma 3.4. By Suzuki [17, p. 417], up to conjugacy the
maximal subgroups of ${\rm PSL}(2,p)$ are isomorphic to:

(i) a dihedral group of order $p+1$ or $p-1$;

(ii) the image of $U$ in ${\rm PSL}(2,p)$;

(iii) the alternating group $A_{4}$;

(iv) the symmetric group $\Sigma_{4}$; or

(v) the alternating group $A_{5}$.

From Table 4, recall that ${\rm PSL}(2,p)$ has an irreducible character of
degree $p+1$. We use Lemma 3.3 to show that the subgroups (conjugate to) (i),
(iii), (iv), and (v) do not form strong Gelfand pairs with ${\rm PSL}(2,p)$.
We will then show (ii) does form a strong Gelfand pair with ${\rm PSL}(2,p)$,
and consider its maximal subgroups.

Case (i): Suppose $(p+1)/2$ is odd. The total character of the dihedral group
of order $p+1$ has degree $2+2((p+1)/2-1)/2$ = $1+(p+1)/2<p+1$ for all primes.
Thus by Lemma 3.3, this dihedral group does not form a strong Gelfand pair
with ${\rm PSL}(2,p)$.

Now suppose $(p+1)/2$ is even. The total character of the dihedral group of
order $p+1$ has degree $4+2((p+1)/4-1)=2+(p+1)/2<p+1$ for all $p>5$. Thus,
Lemma 3.3 again gives the result.

On the other hand, suppose $(p-1)/2$ is odd. The total character of the
dihedral group of order $p-1$ has degree $2+2((p-1)/2-1)/2=1+(p-1)/2<p+1$ for
all primes. Thus, Lemma 3.3 gives the result for this case.

Finally, suppose $(p-1)/2$ is even. The total character of the dihedral group
of order $p-1$ has degree $4+2((p-1)/4-1)=2+(p-1)/2<p+1$ for all primes. The
result follows from Lemma 3.3.

Case (iii): The total character of $A_{4}$ has degree $6$, which is less than
$p+1$ for all $p>5$. Thus, by Lemma 3.3, this case does not produce a strong
Gelfand pair.

Case (iv): Similarly, the total character of $\Sigma_{4}$ has degree $10$,
which is less than $p+1$ for all $p>7$. It follows from Lemma 3.3 that $(\rm
PSL(2,p),\Sigma_{4})$ is not a strong Gelfand pair.

Case (v): Finally, the total character of $A_{5}$ has degree $16$ which is
less than $p+1$ for all primes $p>13$. For the case $p=13$, we see that $60$
does not divide the order of ${\rm SL}(2,13)$. So $A_{5}$ is not a subgroup of
${\rm SL}(2,13)$. Thus, Lemma 3.3 gives the result.

We have now shown that if a subgroup $H$ of ${\rm SL}(2,p)$ forms a strong
Gelfand pair, $H$ must be a subgroup of $U$. To this end, we next show that
$(G,U)$ is a strong Gelfand pair. This is given by the following result.

######  Proposition 4.2.

We have the following decompositions of the characters of $G$ restricted to
$U$:

(i) $(1_{G})|_{U}=1_{U}$;

(ii) $\varphi|_{U}=1_{U}+\chi_{1,2}+\chi_{2,2}$;

(iii) if $i$ is odd,
$\chi_{i}|_{U}=\chi_{0,i}+\chi_{0,(p-1-i)}+\chi_{1,1}+\chi_{2,1}$;

(iv) if $i$ is even,
$\chi_{i}|_{U}=\chi_{0,i}+\chi_{0,(p-1-i)}+\chi_{1,2}+\chi_{2,2}$;

(v) if $j$ is odd, $\theta_{j}|_{U}=\chi_{1,1}+\chi_{2,1}$;

(vi) if $j$ is even, $\theta_{j}|_{U}=\chi_{1,2}+\chi_{2,2}$;

(vii) $\xi_{1}|_{U}=\chi_{0,(p-1)/2}+\chi_{1,2}$;

(viii) $\xi_{2}|_{U}=\chi_{0,(p-1)/2}+\chi_{2,2}$;

(ix) $\eta_{1}|_{U}=\chi_{1,1}$;

(x) $\eta_{2}|_{U}=\chi_{2,1}$;

(xi) $\xi_{1}^{\prime}|_{U}=\chi_{0,(p-1)/2}+\chi_{1,1}$;

(xii) $\xi_{2}^{\prime}|_{U}=\chi_{0,(p-1)/2}+\chi_{2,1}$;

(xiii) $\eta_{1}^{\prime}|_{U}=\chi_{1,2}$;

(xiv) $\eta_{2}^{\prime}|_{U}=\chi_{2,2}$.

###### Proof.

We note that each side of any of the above equations is a character. Thus to
show that the two sides are equal, we take the sum of the characters given on
the right and show this equals the restriction given on the left.

We will frequently use the fact that $Z+Z^{(t)}=-1$ (see Lemma 2.1).

(i): Clearly the restriction of $1_{G}$ to $U$ gives $1_{U}$.

(ii): This case follows from the calculation below:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$1_{U}$ | 1 | 1 | 1 | 1 | 1 | 1 | 1
$\chi_{1,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$
$\chi_{2,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$
$\varphi|_{U}$ | $p$ | $p$ | 1 | 0 | 0 | 0 | 0

(iii): Here $i$ and $p$ are both odd. Thus, $(-1)^{i}=-1$ and
$(-1)^{p-1-i}=-1$. Additionally,
$\zeta^{(p-i-1)j}=\zeta^{(p-1)j}\zeta^{-ij}=\zeta^{-ij}$. This case follows
from:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,i}$ | $1$ | $-1$ | $\zeta^{ij}$ | 1 | 1 | $-1$ | $-1$
$\chi_{0,(p-1-i)}$ | $1$ | $-1$ | $\zeta^{-ij}$ | 1 | 1 | $-1$ | $-1$
$\chi_{1,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$
$\chi_{2,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$
$\chi_{i}|_{U}$ | $p+1$ | $-(p+1)$ | $\zeta^{ij}+\zeta^{-ij}$ | 1 | 1 | -1 | -1

(iv): Here $i$ is even and $p$ is odd. Thus, $(-1)^{i}=1$ and
$(-1)^{p-1-i}=1$. In this case, we have:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,i}$ | $1$ | $1$ | $\zeta^{ij}$ | 1 | 1 | $1$ | $1$
$\chi_{0,(p-1-i)}$ | $1$ | $1$ | $\zeta^{-ij}$ | 1 | 1 | $1$ | $1$
$\chi_{1,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$
$\chi_{2,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$
$\chi_{i}|_{U}$ | $p+1$ | $p+1$ | $\zeta^{ij}+\zeta^{-ij}$ | 1 | 1 | 1 | 1

(v): Here $j$ is odd, and we get:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{1,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$
$\chi_{2,1}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$
$\theta_{j}|_{U}$ | $p-1$ | $-(p-1)$ | 0 | -1 | -1 | 1 | 1

(vi): Here $j$ is even, and:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{1,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$
$\chi_{2,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$
$\theta_{j}|_{U}$ | $p-1$ | $p-1$ | 0 | -1 | -1 | -1 | -1

(vii): Because $p\equiv 1\operatorname{mod}4$, we get $(-1)^{(p-1)/2}=1$. We
also note that $\zeta^{(p-1)j/2}=(\zeta^{(p-1)/2})^{j}=(-1)^{j}$. Then we
have:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,(p-1)/2}$ | 1 | 1 | $(-1)^{j}$ | 1 | 1 | 1 | $1$
$\chi_{1,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$
$\xi_{1}|_{U}$ | $(p+1)/2$ | $(p+1)/2$ | $(-1)^{j}$ | $-Z^{(t)}$ | $-Z$ | $-Z^{(t)}$ | $-Z$

(viii): This case gives:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,(p-1)/2}$ | 1 | 1 | $(-1)^{j}$ | 1 | 1 | 1 | 1
$\chi_{2,2}$ | $(p-1)/2$ | $(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$
$\xi_{2}|_{U}$ | $(p+1)/2$ | $(p+1)/2$ | $(-1)^{j}$ | $-Z$ | $-Z^{(t)}$ | $-Z$ | $-Z^{(t)}$

(ix): For this case, we simply have:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{1,1}=\eta_{1}|_{U}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$

(x): It is seen that:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{2,1}=\eta_{2}|_{U}$ | $(p-1)/2$ | $-(p-1)/2$ | 0 | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$

(xi): Because $p\equiv 3\operatorname{mod}4$, we get $(-1)^{(p-1)/2}=-1$. Then
we compute:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,(p-1)/2}$ | 1 | $-1$ | $(-1)^{j}$ | 1 | 1 | 1 | $1$
$\chi_{1,1}$ | $(p-1)/2$ | $-(p-1)/2$ | $0$ | $Z$ | $Z^{(t)}$ | $-Z$ | $-Z^{(t)}$
$\xi_{1}^{\prime}|_{U}$ | $(p+1)/2$ | $-(p+1)/2$ | $(-1)^{j}$ | $-Z^{(t)}$ | $-Z$ | $Z^{(t)}$ | $Z$

(xii): For this case, we do the following:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{0,(p-1)/2}$ | 1 | $-1$ | $(-1)^{j}$ | 1 | 1 | 1 | 1
$\chi_{2,1}$ | $(p-1)/2$ | $-(p-1)/2$ | $0$ | $Z^{(t)}$ | $Z$ | $-Z^{(t)}$ | $-Z$
$\xi_{2}^{\prime}|_{U}$ | $(p+1)/2$ | $-(p+1)/2$ | $(-1)^{j}$ | $-Z$ | $-Z^{(t)}$ | $Z$ | $Z^{(t)}$

(xiii): We calculate the following restriction:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{1,2}=\eta_{1}^{\prime}|_{U}$ | $(p-1)/2$ | $(p-1)/2$ | 0 | $Z$ | $Z^{(t)}$ | $Z$ | $Z^{(t)}$

(xiv): Finally, we see:

| 1 | $-1$ | $a^{j}$ | $b$ | $d$ | $-b^{-1}$ | $-d^{-1}$
---|---|---|---|---|---|---|---
$\chi_{2,2}=\eta_{2}^{\prime}|_{U}$ | $(p-1)/2$ | $(p-1)/2$ | 0 | $Z^{(t)}$ | $Z$ | $Z^{(t)}$ | $Z$

This completes consideration of all cases, and we conclude that $(G,U)$ is a
strong Gelfand pair. ∎

Because $(G,U)$ is a strong Gelfand pair, we must now consider the maximal
subgroups of $U$ to determine if any of them forms a strong Gelfand pair with
$G$.

######  Proposition 4.3.

For any proper subgroup $K\leq U$, $K\neq H_{2}$, the pair $(G,K)$ is not a
strong Gelfand pair.

When $p\equiv 1\operatorname{mod}4$, the pair $(G,H_{2})$ is not a strong
Gelfand pair.

When $p\equiv 3\operatorname{mod}4$, the pair $(G,H_{2})$ is a strong Gelfand
pair.

###### Proof.

Recall that $U=\langle a,b\rangle=\langle a\rangle\rtimes\langle b\rangle$, so
for any subgroup $K\leq U$ we have the following possibilities:

(i) $p\nmid|K|$, in which case $K$ is conjugate to $\langle a^{q}\rangle$ for
some $q$ dividing $p-1$. The maximal subgroups of $U$ containing $K$ are all
conjugate to $A=\langle a\rangle$, the cyclic group of order $p-1$. We see
that

$\langle\varphi|_{A},1_{A}\rangle=\frac{1}{p-1}(p+p+1+\cdot\cdot\cdot+1)=3.$

Thus, $(G,A)$ is not a strong Gelfand pair.

(ii) $p\big{|}|K|$, in which case $K=\langle a^{q}\rangle\rtimes\langle
b\rangle$ for some $q$ dividing $p-1$. The maximal subgroups are those with
$q$ prime. We define $H_{q}=\langle a^{q}\rangle\rtimes\langle b\rangle$,
which has index $q$ in $U$. Note that this corresponds with our earlier
definition of $H_{2}$. Let $p_{q}=(p-1)/q$.

First, we look at $q>2$. Then $p_{q}\leq(p-1)/3<(p-3)/2$, so referring to the
bounds for Table 2 we consider the character $\chi_{p_{q}}$ of $G$. We find
that the inner product $\langle\chi_{p_{q}}|_{H_{q}},1_{H_{q}}\rangle$ is

$\displaystyle\frac{1}{p\cdot p_{q}}\Big{(}(p+1)1$
$\displaystyle+(p+1)1+1\cdot p_{2}+1\cdot p_{2}+1\cdot p_{2}+1\cdot
p_{2}+2p(p_{q}-2)\Big{)}$
$\displaystyle=\frac{q}{p(p-1)}\left(4p+2p\left(\frac{p-1}{q}\right)-4p\right)=2.$

Thus, $(G,H_{q})$ is not a strong Gelfand pair whenever $q>2$.

When $q=2$, there are two cases. First, consider $p\equiv
1\operatorname{mod}4$. As we can deduce from Table 1, each linear character of
$H_{2}$ is the restriction of two different linear characters of $U$. In
particular, $\chi_{0,(p-1)/4}|_{H_{2}}=\chi_{0,3(p-1)/4}|_{H_{2}}$. From case
(iv) of Proposition 4.2,

$\displaystyle\chi_{(p-1)/4}|_{H_{2}}$
$\displaystyle=\chi_{0,(p-1)/4}|_{H_{2}}+\chi_{0,3(p-1)/4}|_{H_{2}}+\chi_{1,2}|_{H_{2}}+\chi_{2,2}|_{H_{2}}$
$\displaystyle=2\chi_{0,(p-1)/4}|_{H_{2}}+\chi_{1,2}|_{H_{2}}+\chi_{2,2}|_{H_{2}},$

and we conclude that $(G,H_{2})$ is not a strong Gelfand pair.

Finally, if $p\equiv 3\operatorname{mod}4$, we show that $(G,H_{2})$ is a
strong Gelfand pair. Because the subgroup $H_{2}$ has index 2 in $H$, the
result in [12, Prop. 20.10] applies. Since $p\equiv 3\operatorname{mod}4$,
$|H_{2}|=p(p-1)/2$ is odd, $-1\notin H_{2}$. Thus, because $\chi(-1)\neq 0$
for all $\chi\in\hat{H}$, we have that each irreducible character of $H_{2}$
is simply the restriction of an irreducible character of $H$. It follows by
Proposition 4.2 that $(G,H_{2})$ is a strong Gelfand pair.

Because $(G,H_{2})$ is a strong Gelfand pair when $p\equiv
3\operatorname{mod}4$, we must now consider subgroups of $H_{2}$. We show that
no proper subgroup of $H_{2}$ forms a strong Gelfand pair with $G$. Because
$H_{2}=\langle a^{2}\rangle\rtimes\langle b\rangle$, any subgroup of $H_{2}$
is of the form (i) $\langle a^{2k}\rangle\rtimes\langle b\rangle$ or (ii)
$\langle a^{2k}\rangle$. The subgroups in (i) will be maximal when $k$ is
prime; the maximal subgroup in (ii) is $\langle a^{2}\rangle$.

(i) If $k$ is an odd prime, we see that $\langle a^{2k}\rangle\rtimes\langle
b\rangle$ will be a subgroup of $H_{k}$. Because $(G,H_{k})$ is not a strong
Gelfand pair, it follows that $\langle a^{2k}\rangle\rtimes\langle b\rangle$
also does not form a strong Gelfand pair with $G$.

If $k=2$, then $4|p-1$, and we have $H_{4}=\langle a^{4}\rangle\rtimes\langle
b\rangle$. Then $(p-1)/4<(p-3)/2$, so referring to the bounds for Table 2 we
consider the character $\chi_{p_{4}}$ of $G$. We find that the inner product
$\langle\chi_{p_{4}}|_{H_{4}},1_{H_{4}}\rangle$ is

$\displaystyle\frac{1}{p\cdot p_{4}}\Big{(}(p+1)1$
$\displaystyle+(p+1)1+1\cdot p_{2}+1\cdot p_{2}+1\cdot p_{2}+1\cdot
p_{2}+2p(p_{4}-2)\Big{)}$
$\displaystyle=\frac{4}{p(p-1)}\left(4p+2p\left(\frac{p-1}{4}\right)-4p\right)=2.$

So $(G,H_{4})$ is not a strong Gelfand pair.

(ii) We see that $\langle a^{2}\rangle$ is a subgroup of $A$. Thus, by Lemma
3.1 and because $(G,A)$ is not a strong Gelfand pair, $(G,\langle
a^{2}\rangle)$ is also not a strong Gelfand pair.

Thus, we conclude that there is no subgroup $K$ of $H_{2}$ such that $(G,K)$
forms a strong Gelfand pair.

Applying Lemma 3.1, we have shown that the only proper subgroup of $U$ which
forms a strong Gelfand pair with $G$ is the subgroup $H_{2}$ when $p\equiv
3\operatorname{mod}4$. ∎

The above applies whenever $p\geq 13$, and so concludes the proof of Theorem
1.1 and Theorem 4.1. ∎

To finish, we consider individually the cases $p=2,3,5,7,\text{ and }11$:

######  Theorem 4.4.

(i) $p=2$. In this case, $G\cong\Sigma_{3}$, and the strong Gelfand pairs are
$(G,\mathcal{C}_{2})$ and $(G,\mathcal{C}_{3})$.

(ii) $p=3$. Then the strong Gelfand pairs are $(G,\mathcal{C}_{3})$,
$(G,\mathcal{C}_{6})$, and $(G,U)$.

(iii) $p=5$. Then the strong Gelfand pairs are $(G,{\rm SL}(2,3))$ and
$(G,U)$, where ${\rm SL}(2,3)\leq SL(2,5)$ is maximal.

(iv) $p=7$. Then the strong Gelfand pairs are $(G,U)$, $(G,H_{2})$, and
$(G,K)$, where $K$ is the preimage of $\Sigma_{4}$ in ${\rm PSL}(2,7)$. There
are two non-conjugate copies of $K$ in $G$.

(v) $p=11$. Then the strong Gelfand pairs are $(G,H_{2})$, $(G,U)$, and
$(G,2{\rm I)}$, where

$2{\rm I}=\langle r,s,t|r^{2}=s^{3}=t^{5}=rst\rangle$

is the binary icosahedral group. There are two non-conjugate copies of $2{\rm
I}$ in $G$.

###### Proof.

These calculations were made using Magma [18]. We give generating matrices for
those subgroups which have not been defined previously:

(iii): ${\rm SL}(2,3)$ as a subgroup of ${\rm SL}(2,5)$ is generated by

$\begin{pmatrix}3&3\\\ 4&1\end{pmatrix},\begin{pmatrix}4&1\\\
3&1\end{pmatrix},\begin{pmatrix}3&0\\\ 1&2\end{pmatrix},\begin{pmatrix}4&0\\\
0&4\end{pmatrix}.$

(iv): One copy of $K$ is generated by

$\begin{pmatrix}4&2\\\ 2&3\end{pmatrix},\begin{pmatrix}2&0\\\
6&4\end{pmatrix},\begin{pmatrix}3&2\\\ 2&4\end{pmatrix},\begin{pmatrix}0&6\\\
1&0\end{pmatrix},\begin{pmatrix}6&0\\\ 0&6\end{pmatrix};$

the other is generated by

$\begin{pmatrix}3&6\\\ 3&4\end{pmatrix},\begin{pmatrix}5&0\\\
3&3\end{pmatrix},\begin{pmatrix}2&6\\\ 5&5\end{pmatrix},\begin{pmatrix}4&4\\\
1&3\end{pmatrix},\begin{pmatrix}6&0\\\ 0&6\end{pmatrix}.$

(v): The first copy of $2{\rm I}$ is generated by

$\begin{pmatrix}7&2\\\ 8&4\end{pmatrix},\begin{pmatrix}6&2\\\
1&6\end{pmatrix},\begin{pmatrix}10&0\\\ 0&10\end{pmatrix};$

the other is generated by

$\begin{pmatrix}6&9\\\ 2&5\end{pmatrix},\begin{pmatrix}8&5\\\
4&4\end{pmatrix},\begin{pmatrix}10&0\\\ 0&10\end{pmatrix}.$

These subgroups are also described in [8]. ∎

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11institutetext: DEIM, Universitat Rovira i Virgili, Spain. 22institutetext:
Imaging Informatics Division, Bioinformatics Institute, Singapore.
33institutetext: Department of Medicine and Health Sciences, Universitat
Rovira i Virgili, Spain.

# An Efficient Solution for Breast Tumor Segmentation and Classification in
Ultrasound Images Using Deep Adversarial Learning

Vivek Kumar Singh 11Corresponding Author:
<EMAIL_ADDRESS>Author<EMAIL_ADDRESS>Hatem
A. Rashwan 11 Mohamed Abdel-Nasser 11 Md. Mostafa Kamal Sarker 11 Farhan Akram
22 Nidhi Pandey 33 Santiago Romani 11 Domenec Puig 11

###### Abstract

This paper proposes an efficient solution for tumor segmentation and
classification in breast ultrasound (BUS) images. We propose to add an atrous
convolution layer to the conditional generative adversarial network (cGAN)
segmentation model to learn tumor features at different resolutions of BUS
images. To automatically re-balance the relative impact of each of the highest
level encoded features, we also propose to add a channel-wise weighting block
in the network. In addition, the SSIM and L1-norm loss with the typical
adversarial loss are used as a loss function to train the model. Our model
outperforms the state-of-the-art segmentation models in terms of the Dice and
IoU metrics, achieving top scores of $93.76\%$ and $88.82$%, respectively. In
the classification stage, we show that few statistics features extracted from
the shape of the boundaries of the predicted masks can properly discriminate
between benign and malignant tumors with an accuracy of $85\%$.

###### Keywords:

Breast Cancer; Ultrasound Image Segmentation; CAD System; Deep Adversarial
Learning

##  1 Introduction

Breast cancer is one of the most commonly diagnosed causes of death in women
worldwide [14]. Screening with mammography can recognize tumors in the early
stages. Despite, some breast cancers may not be captured in mammographies
(e.g., in the case of dense breasts). Ultrasound has been recommended as a
powerful adjunct screening tool for detecting breast cancers that may be
occluded in mammographies [8]. Computer-aided diagnosis (CAD) systems are
widely used to detect, segment and classify masses in breast ultrasound (BUS)
images. One of the main steps of BUS CAD systems is tumor segmentation.
Over the last two decades, several BUS image segmentation methods have been
proposed, which can be categorized into semi-automated and fully automated
according to the degree of human intervention. In [1], a region growing based
algorithm was used to automatically extract the regions that contain the
tumors, and image super-resolution and texture analysis methods were used to
discriminate benign tumors from the malignant ones. Recently, some deep
learning based models have been proposed to improve the performance of breast
tumor segmentation methods. In [16], two convolutional neural network (CNN)
architectures have been used to segment BUS images into the skin, mass, fibro-
glandular, and fatty tissues (an accuracy of $90$%). Hu et al [5] combined a
dilated fully convolutional network with a phase-based active contour model to
segment breast tumors, achieving dice score of $88.97\%$.
Although these methods and others proposed in the literature do provide useful
techniques, there are still challenges due to the high degree of speckle noise
present in the ultrasound images, as well as to the high variability of tumors
in shape, size, appearance, texture, and location. In this paper, we propose
an efficient solution for breast tumor segmentation and classification in BUS
images using deep adversarial learning.
The main contribution of this paper is to develop an efficient deep model for
segmenting the breast tumor in BUS by combining an atrous convolution network
(AC) and channel attention with channel weighting (CAW) in a cGAN model in
order to enhance the discriminant ability of feature representations at multi-
scale. Besides, we demonstrate that the proposed segmentation model can be
used for characterizing accurate shape features from the segmented mask to
discriminate between benign and malignant tumors. The rest of this paper is
structured as follows. Section 2 presents the proposed model. Section 3
includes the results. Section 4 concludes our study and provides some lines of
future work.

##  2 Proposed Methodology

Generative adversarial network architecture: The proposed BUS image
segmentation technique is based on generative adversarial training, which
involves two interdependent networks: a generator G and a discriminator D.
(Fig. 1). The generator generates a fake example from input noise z, while
discriminator determines the probability that the fake example is from
training data rather than generated by the generator.
Generator: The generator network incorporates an encoder section, made of
seven convolutional layers (En1 to En7), and a decoder section, made of seven
deconvolutional layers (Dn1 to Dn7) layers. We have modified the plain
encoder-decoder structure by inserting an atrous convolution block [18]
between En3 and En4, in addition to a CAW block between En7 and Dn1. The CAW
block is an aggregation of a channel attention module [3] with channel
weighting block [4]. In turn, the CAW block increases the representational
power of the highest level features of the generator network, which turns out
in a clear improvement of the accuracy of the breast tumor segmentation in
ultrasound images (for more details, see suppl. A.1 and A.2).

Figure 1: The architecture of the proposed segmentation model for BUS images.

By including the atrous convolutional block in-between encoder layers En3 and
En4, the generator network is enabled to characterize features at different
scales and also to expand the actual receptive field of the filters. As a
consequence, the network is more aware of contextual information without
increasing the number of parameters or the amount of computation. We use 1, 6
and 9 dilation rates with kernel size $3\times 3$ and a stride of 2.
Each layer in the encoder section is followed by batch normalization (except
for En1 and En7) and LeakyReLU with slope 0.2, except for En7, where the
regular non-linearity ReLU activation function is used. The decoder section is
a sequence of transposed-convolutional layers followed by batch normalization,
dropout with rate $0.5$ (only in Dn1, Dn2, and Dn3) and ReLU. The filters of
the convolutional and deconvolutional layers are defined by a kernel of
$4\times 4$ and they are shifted with a stride of 2. We add padding of 2 after
En4, yielding a $4\times 4\times 512$ output feature map. We also add skip
connection between the corresponding layers in the encoder and decoder
sections, which improve the features in the output image by merging deep,
coarse, semantic information and simple, fine, appearance information. After
the last decoding layer (Dn7), the tanh activation function is used as a non-
linear output of the generator, which is trained to generate a binary mask of
the breast tumor.
Discriminator: It is a sequence of convolutional layers applying kernels of
size $4\times 4$ with a stride of 2, except for Cn4 and Cn5 where the stride
is 1. Batch normalization is employed after Cn2 to Cn4. LeakyReLU with slope
0.2 is the non-linear activation function used after Cn1 to Cn4, while the
sigmoid function is used after Cn5. The input of the discriminator is the
concatenation of the BUS image and a binary mask marking the tumor area, where
the mask can either be the ground truth or the one predicted by the generator
network. The output of the discriminator is a $10\times 10$ matrix having
values varying from 0.0 (completely fake) to 1.0 (real).
Loss Functions: Assume $x$ is a BUS image containing a breast tumor, $y$ is
the ground truth mask of that tumor within the image, $G(x,z)$ and
$D(x,G(x,z))$ are the outputs of the generator and the discriminator,
respectively. The loss function of the generator $G$ comprises three terms:
adversarial loss (binary cross entropy loss), L1-norm to boost the learning
process, and SSIM loss [15] to improve the shape of the boundaries of
segmented masks:

$\begin{split}\ell_{Gen}(G,D)=\mathbb{E}_{x,y,z}(-\log(D(x,G(x,z))))+\\\
\lambda\mathbb{E}_{x,y,z}(\ell_{L1}(y,G(x,z)))+\alpha\mathbb{E}_{x,y,z}(\ell_{SSIM}(y,G(x,z)))\end{split}$
(1)

where $z$ is a random variable and $\lambda$ and $\alpha$ are empirical
weighting factors. The variable $z$ is introduced as a dropout in the decoding
layers $Dn1$, $Dn2$ and $Dn3$ at both training and testing phases, which helps
to generalize the learning processes and avoid overfitting. If the generator
network is properly optimized, the values of $D(x,G(x,z))$ should approach
$1.0$, meaning that discriminator cannot distinguish generated tumor masks
from ground truth masks, while L1 and SSIM losses should approach to $0.0$,
indicating that every generated mask matches the corresponding ground truth
both in overall pixel-to-pixel distances (L1) and in basic statistic
descriptors (SSIM). For more details and analysis of loss functions, the
reader is referred to suppl. A.3.
The loss function of the discriminator $D$ can be formulated as follows:

$\displaystyle\ell_{Dis}(G,D)$
$\displaystyle=\mathbb{E}_{x,y,z}(-\log(D(x,y)))+\mathbb{E}_{x,y,z}(-\log(1-D(x,G(x,z))))$
(2)

The optimizer will fit $D$ to maximize the loss values for ground truth masks
(by minimizing $-\log(D(x,y))$) and minimize the loss values for generated
masks (by minimizing $-\log(1-D(x,G(x,z))$). These two terms compute BCE loss
using both masks, assuming that the expected class for ground truth and
generated masks are $1$ and $0$, respectively. $G$ and $D$ networks are
optimized concurrently: one optimization step for both networks at each
iteration, where $G$ tries to generate a valid tumor segmentation and $D$
learns how to differentiate between the synthetic and real segmentation.
Model training: In the preprocessing step, each BUS images is rescaled to
96x96 pixels, and pixel values are normalized between [0,1]. In the
postprocessing step, morphological operations ($3\times 3$ closing, $2\times
2$ erosion) are used to suppress most of the outlier predictions (speckled
pixels). The hyperparameters of the model were experimentally tuned. We also
explored several optimizers, such as SGD, AdaGrad, Adadelta, RMSProp, and Adam
with different learning rates (see suppl. A.3). We achieved the best results
with Adam optimizer ($\beta_{1}$= 0.5, $\beta_{2}$= 0.999) and learning rate
=0.0002 with a batch size of 8. The SSIM loss and L1-norm loss weighting
factors $\lambda$ and $\alpha$ were set to 10 and 5, respectively. The best
results were achieved by training both generator and discriminator from
scratch for 40 epochs.
Breast Tumor Classification:  To classify the BUS image into benign and
malignant, we propose to rely on statistic features of the segmented tumor
mask to discriminate between both classes. Malignant breast tumors and benign
lesions have different shape characteristics: the malignant lesion usually is
irregular, speculated, or microlobulated. However, benign lesion mainly has
smooth boundaries, round, oval, or macrolobulated shape [17].
In the classification method, each BUS image is fed into the trained
generative network to obtain the boundary of the tumor, and then we compute 13
statistical features from that boundary: fractal dimension, lacunarity, convex
hull, convexity, circularity, area, perimeter, centroid, minor and major axis
length, smoothness, Hu moments (6) and central moments (order 3 and below). We
implemented an Exhaustive Feature Selection (EFS) algorithm to select the best
set of features. The EFS algorithm indicates that the fractal dimension,
lacunarity, convex hull, and centroid are the 4 optimal features. The selected
features are fed into a Random Forest classifier, which is later trained to
discriminate between benign and malignant tumors.

##  3 Experiments and Discussion

BUS dataset. We evaluated the performance of the proposed model using the
Mendeley Data BUS dataset, which is publicly available [11]. This dataset
contains 150 malignant and 100 benign tumors contained in BUS images. To train
our model, we randomly divided the dataset into the training set ($70\%$), a
validation set ($10\%$) and testing set ($20\%$). The dataset does not have a
ground truth for tumor segmentation. Thus, cooperative experts have manually
segmented the tumors appearing in the BUS images.
Data augmentation: To augment the current set of available examples, we
applied different operations: 1) scale the images by factors varying from 0.5
to 2 with steps of 0.25, 2) apply gamma correction on the BUS images by
factors varying from 0.5 to 2.5 with steps of 0.5, and 3) flip and rotate the
images. These operations yield 8K BUS images.
Breast tumor segmentation results: In Table 1, we compare the baseline cGAN
model [6] with three variations of our model: cGAN with atrous convolution
(cGAN+AC), cGAN with channel attention and weighting (cGAN+CAW), and cGAN with
AC and CAW (cGAN+AC+CAW). All of these variations are also compared with six
state-of-the-art image segmentation methods: FCN [10], UNet [13] SegNet[2],
ERFNet [12], and DCGAN [7]. All methods are evaluated both quantitatively and
qualitatively. For the quantitative analysis, we calculate the accuracy (ACC),
Dice (DIC), Intersection over Union (IoU), sensitivity (SEN) and specificity
(SPE) metrics. As shown in Table 1, the added AC and CAW blocks improves the
results of the baseline cGAN model. In addition, our model (cGAN+AC+CAW)
outperforms the rest in all metrics. It achieves Dice and IoU scores of
$93.76\%$ and $88.82\%$, respectively, which are the metrics that better
represent the degree of coincidence between predicted and ground truth
segmentation. These two results outperform the ones from the second best model
in the table, the UNet model, in $5\%$ to $6\%$ absolute points over the full
range, which is quite significant taking into account their proximity to the
maximum value. The SegNet and ERFNet models yield the worst segmentation
results on BUS images. The results of the proposed model have also been
compared with other methods evaluated on different datasets. For instance, Hu
et al [5] yielded an IoU of $85.10$% on a private BUS image dataset. In
addition, Xu et al [16] achieved a Dice score of $89.00$%.

Table 1: Segmentation results of the proposed model(cGAN+AC+CAW) and compared models FCN [10], SegNet[2], UNet [13], ERFNet [12], DCGAN [7] and cGAN[6].  | Methods | DIC% | IoU%
---|---|---|---
| FCN |  $79.73\pm 0.102$ | $66.29\pm 0.116$
SegNet |  $50.35\pm 0.295$ | $41.65\pm 0.274$
UNet |  $88.28\pm 0.090$ | $82.23\pm 0.096$
ERFNet |  $67.02\pm 0.206$ | $53.32\pm 0.185$
DCGAN |  $85.55\pm 0.154$ | $75.28\pm 0.128$
cGAN |  $86.04\pm 0.095$ | $77.56\pm 0.080$
cGAN + AC |  $87.14\pm 0.084$ | $80.77\pm 0.081$
cGAN + CAW |  $90.65\pm 0.075$ | $83.40\pm 0.069$
| Ours |  ${\color[rgb]{1,0,0}\textbf{93.76}\pm 0.037}$ | ${\color[rgb]{1,0,0}\textbf{88.82}\pm 0.064}$

Figure 2 shows boxplots of Dice and IoU values obtained for the 50 testing
samples using FCN, SegNet, ERFNet, UNet and the proposed model. The two models
based on cGAN provided small ranges of Dice and IoU values. For instance, our
model is in the range $88$% to $94$% for Dice coefficient and $80$% to $89$%
for IoU, while other deep segmentation methods, FCN, SegNet, ERFNet and UNet
show a wider range of values. Moreover, there are many outliers in the results
for the segmentation based on the other baseline segmentation methods, while
using our proposal there are none.

Figure 2: Boxplots of IoU and Dice metrics of the proposed model and FCN [10],
SegNet[2], ERFNet [12] and UNet [13].

Figure 3 presents a comparison between image segmentation obtained with our
model and the other six models. As shown, SegNet and ERFNet yield the worst
results since there are large false negative areas (in red), as well as some
false positive areas (in green). FCN also shows rather significant erroneous
areas, although it has fairly segmented the example (b). In turn, UNet, DCGAN,
cGAN provide good segmentation but our model provide more accurate
segmentation of the boundary of breast tumors, although our model shows the
smallest erroneous areas, from a visual inspection.

Figure 3: Segmentation results on four samples of the BUS dataset. The rows
(a) and (b) show benign samples while rows (c) and (d) rows show malignant
samples.

Breast tumor classification results: To test our classification strategy, we
have checked our method with different segmentation method output with the
leave-one-out cross-validation technique and calculated the precision, recall,
accuracy and F1-score metrics. Furthermore, we have also obtained the same
metrics from the work of Lee et.al [9], who proposed a stack denoising
autoencoder method to segment and classify breast tumors from the same BUS
dataset that we use in this study. As shown in Table 2, the proposed breast
tumor classification method outperforms [9], with a total accuracy degree of
$85\%$.

Table 2: Breast tumor classification results Methods | Precision | Recall | Accuracy | Measure
---|---|---|---|---
FCN | $62.0\pm 0.138$ |  $70.0\pm 0.154$ | $69.0\pm 0.128$ |  $68.0\pm 0.135$
SegNet | $51.0\pm 0.216$ |  $62.0\pm 0.238$ | $53.0\pm 0.200$ |  $51.0\pm 0.229$
UNet | $70.0\pm 0.098$ |  $81.0\pm 0.104$ | $77.0\pm 0.091$ |  $77.0\pm 0.087$
ERFNet | $58.0\pm 0.162$ |  $66.0\pm 0.205$ | $61.0\pm 0.146$ |  $59.0\pm 0.172$
DCGAN | $71.0\pm 0.098$ |  $82.0\pm 0.081$ | $75.0\pm 0.105$ |  $73.0\pm 0.107$
cGAN | $71.0\pm 0.083$ |  $84.0\pm 0.095$ | $78.0\pm 0.079$ |  $77.0\pm 0.776$
cGAN+AC | $73.0\pm 0.060$ |  $87.0\pm 0.068$ | $80.0\pm 0.053$ |  $81.0\pm 0.059$
cGAN+CAW | $74.0\pm 0.052$ |  $88.0\pm 0.070$ | $82.0\pm 0.056$ |  $82.0\pm 0.054$
Lee et. al [9] | $78.0$ |  $90.0$ | $83.0$ |  $83.0$
Ours | ${\color[rgb]{1,0,0}\textbf{81.0}\pm 0.021}$ |  ${\color[rgb]{1,0,0}\textbf{92.0}\pm 0.028}$ | ${\color[rgb]{1,0,0}\textbf{85.0}\pm 0.196}$ |  ${\color[rgb]{1,0,0}\textbf{84.0}\pm 0.024}$

##  4 Conclusion

In this paper, we have proposed an efficient solution for tumor segmentation
and classification in BUS images. We have proposed to add an atrous
convolution blocks to the generator network to learn tumor features at
different resolutions of BUS images. We also have used a channel-wise
weighting block in the generator network to automatically re-balance the
relative impact of each of the highest level encoded features. Our model
outperforms the FCN, SegNet, ERFNet, UNet, DCGAN and cGAN segmentation models
in terms of Dice and IoU metrics, achieving the top scores of $93.76\%$ and
$88.82\%$ respectively. In the classification stage, we used four optimal
statistics features extracted from the segmented tumor masks, obtaining an
accuracy of $85\%$, which is $2\%$ over related method that uses the same
database. As future work, we will focus on adapting the proposed model to
segment lesions of different organs using multi-modal medical images.

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