# On the automorphism groups of Lunardon-Polverino scattered linear sets

Wei Tang 1 ,  Yue Zhou 1,$\dagger$ and  Ferdinando Zullo 2 1Department of
Mathematics, National University of Defense Technology, 410073 Changsha, China
<EMAIL_ADDRESS>2Department of Mathematics and Physics, University of
Campania “Luigi Vanvitelli”, 81012 Caserta, Italy
<EMAIL_ADDRESS>$\dagger$Corresponding author

###### Abstract.

Lunardon and Polverino introduced in 2001 a new family of maximum scattered
linear sets in $\mathrm{PG}(1,q^{n})$ to construct linear minimal Rédei
blocking sets. This family has been extended first by Lavrauw, Marino,
Trombetti and Polverino in 2015 and then by Sheekey in 2016 in two different
contexts (semifields and rank metric codes). These linear sets are called
_Lunardon-Polverino_ linear sets and this paper aims to determine their
automorphism groups, to solve the equivalence issue among Lunardon-Polverino
linear sets and to establish the number of inequivalent linear sets of this
family. We then elaborate on this number, providing explicit bounds and
determining its asymptotics.

###### Key words and phrases:

linear set, rank-metric code, automorphism group

##  1\. Introduction

The determination of the automorphism group of combinatorial objects has been
a central topic in combinatorics for many years, especially because it gives
information on the _symmetry_ of the considered structure. However, very
little is known about the automorphism groups of linear sets.

Let $q=p^{r}$, for some prime $p$ and a positive integer $r$. Let $n$ be a
positive integer and let $V$ be a $2$-dimensional vector space over
$\mathbb{F}_{q^{n}}$ (up to coordinatize, we can suppose that
$V=\mathbb{F}_{q^{n}}^{2}$). A point set $L$ of
$\Lambda=\mathrm{PG}(V,\mathbb{F}_{q^{n}})=\mathrm{PG}(1,q^{n})$ is said to be
an _$\mathbb{F}_{q}$ -linear set_ of $\Lambda$ of rank $k$ if it is defined by
the non-zero vectors of a $k$-dimensional $\mathbb{F}_{q}$-vector subspace $U$
of $V$, i.e.

$L=L_{U}:=\\{\langle{\bf u}\rangle_{\mathbb{F}_{q^{n}}}:{\bf u}\in
U\setminus\\{{\bf 0}\\}\\}.$

For any point $P=\mathrm{PG}(Z,\mathbb{F}_{q^{n}})$ of $\Lambda$, the _weight_
of $P$ in $L_{U}$ is defined as $w_{L_{U}}(P)=\dim_{\mathbb{F}_{q}}(U\cap Z)$.
For an $\mathbb{F}_{q}$-linear set $L$ of rank $k$ the following holds

(1) $|L|\leq\frac{q^{k}-1}{q-1}.$

We also recall that two linear sets $L_{U}$ and $L_{W}$ in the projective line
$\Lambda$ are said to be _$\mathrm{P\Gamma L}$ -equivalent_ (or simply
_equivalent_) if there is an element $\varphi$ in $\mathrm{P\Gamma
L}(2,q^{n})$ such that $L_{U}^{\varphi}=L_{W}$. In general, it is very
challenging to determine whether or not two linear sets are equivalent; see
e.g. [7].

An important family of linear sets are the _scattered_ ones, that is those
linear sets satisfying the equality in (1) (or, equivalently, the linear sets
whose points have weight one); see e.g. [20]. Scattered linear sets were
originally introduced by Blokhuis and Lavrauw in [6] in a more general
setting. Since the seminal paper by Blokhuis and Lavrauw, linear sets
attracted a lot of attention especially because of their connections with
blocking sets, semifields, rank metric codes, etc (see e.g. [13, 20, 21]). In
[6], it has also been proved that a scattered linear set in $\Lambda$ can have
rank at most $n$ and a scattered linear set of rank $n$ is said to be _maximum
scattered_. The first example of maximum scattered linear set is the linear
set _of pseudoregulus type_ , that is

$\\{\langle(x,x^{q^{s}})\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\\},$

with $\gcd(s,n)=1$. In particular,

$\left\\{\langle(x,x^{q})\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{\langle(x,x^{q^{s}})\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\right\\},$

for any $s$ such that $\gcd(s,n)=1$. This family has been studied in details,
indeed its geometric characterization has been provided by Csajbók and Zanella
in [8].

The next example of maximum scattered linear set in $\Lambda$ which has been
given by Lunardon and Polverino in [16] and generalized in [12, 21] is

$L=\\{\langle(x,x^{q^{s}}+\delta
x^{q^{n-s}})\rangle_{\mathbb{F}_{q^{n}}}\colon x\in\mathbb{F}_{q^{n}}^{*}\\},$

with $n\geq 4$, $\gcd(s,n)=1$, $q\neq 2$, and nonzero $\delta$ such that
$N_{q^{n}/q}(\delta)=\delta^{\frac{q^{n}-1}{q-1}}\neq 1$. This example is
known as _Lunardon-Polverino_ linear set. If we set $\delta=0$, then the
Lunardon-Polverino linear set reduces to be of pseudoregulus type. In [22,
Theorem 3.4], Zanella proved that the condition $N_{q^{n}/q}(\delta)\neq 1$ is
necessary for $L$ to be scattered (see also [2]). Also, a more geometric
description for such linear sets has been given in [23] and an upper bound on
the number of inequivalent Lunardon-Polverino linear sets has been provided in
[15, Proposition 2.3], but the exact number is unknown.

Let $L_{U}$ be an $\mathbb{F}_{q}$-linear set of rank $n$ in $\Lambda$. It is
well-known that ${\rm P\Gamma L}(2,q^{n})$ is $3$-transitive on $\Lambda$, and
so we may assume that $L_{U}$ does not contain the point
$\langle(0,1)\rangle_{\mathbb{F}_{q^{n}}}$, so that $L_{U}$ can be written as

$L_{f}=\\{\langle(x,f(x))\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\\},$

where $f$ is a _$q$ -polynomial_ over $\mathbb{F}_{q^{n}}$, that is
$f=\sum_{i=0}^{n-1}a_{i}X^{q^{i}}\in\mathbb{F}_{q^{n}}[X]$. Sheekey in [21]
called $f$ _scattered_ if $L_{f}$ turns out to be a scattered
$\mathbb{F}_{q}$-linear set. When choosing $f=X^{q^{s}}$ then $L_{f}$ turns
out to be the linear set of pseudoregulus type, whereas if $f=X^{q^{s}}+\delta
X^{q^{n-s}}$ then $L_{f}$ is the Lunardon-Polverino linear set. These two
families of polynomials play an important role in the classifications of
scattered polynomials which are _exceptional_ , that is, roughly speaking,
polynomials which are scattered for infinitely many $n$. Indeed, the known
classification results suggest that these two families are the only examples
of exceptional scattered polynomials; see [4, 3, 5, 9].

In this paper we study Lunardon-Polverino linear sets. We first completely
solve the equivalence issue among two Lunardon-Polverino linear sets and we
also determine the automorphism group of a Lunardon-Polverino linear set. This
has been done by exploiting the problem of determining the relations between
the coefficients of two $q$-polynomials $f$ and $g$ for which the value sets
of $f(x)/x$ and $g(x)/x$ coincide. As a byproduct and making use of careful
computations, we completely determine $\Lambda(n,q)$ the number of
inequivalent Lunardon-Polverino linear sets. Finally we compute explicit upper
and lower bounds on $\Lambda(n,q)$ and we determine its asymptotic behavior as
$r\to+\infty$, by making use of a result by Gronwall (1913) involving the
Euler-Mascheroni constant.

##  2\. Auxiliary results

We need the following well known result on the greatest common divisors. A
proof can be found in [19].

######  Lemma 2.1.

Let $v(b):=\max\\{i:2^{i}\mid b\\}$. For a prime $p$,

(2) $\gcd(p^{i}+1,p^{j}-1)=\begin{cases}p^{\gcd(i,j)}+1,&v(i)<v(j);\\\
\frac{(-1)^{p+1}+3}{2},&\text{otherwise}.\end{cases}$

Let $s$ be a positive integer such that $\gcd(s,n)=1$ and let $\sigma\colon
x\in\mathbb{F}_{q^{n}}\mapsto x^{q^{s}}\in\mathbb{F}_{q^{n}}$. A
$\sigma$-_polynomial_ (or a _linearized polynomial_) over $\mathbb{F}_{q^{n}}$
is a polynomial of the following form

$f=\sum_{i=0}^{\ell}a_{i}X^{\sigma^{i}},$

where $a_{i}\in\mathbb{F}_{q^{n}}$ and $\ell$ is a positive integer. We will
also call $f$ a $q^{s}$-_polynomial_. If $a_{\ell}\neq 0$ we say that $\ell$
is the $\sigma$-_degree_ of $f$. We will denote by $\mathcal{L}_{n,\sigma}$
the set of all $\sigma$-polynomials over $\mathbb{F}_{q^{n}}$ with
$\sigma$-degree less than $n$ (when $s=1$, we will denote it by
$\mathcal{L}_{n,q}$). The set $\mathcal{L}_{n,\sigma}$ equipped with the
classical sum of polynomials, the composition modulo $X^{\sigma^{n}}-X$ and
the scalar multiplication by an element in $\mathbb{F}_{q}$, is an
$\mathbb{F}_{q}$-algebra isomorphic to the algebra of $\mathbb{F}_{q}$-linear
endomorphism of $\mathbb{F}_{q^{n}}$. This isomorphism allows us to identify a
$\sigma$-polynomial with the $\mathbb{F}_{q}$-linear map it defines. We also
want to underline that different choices of $\sigma$ give a different
representation of an element, but leave unchanged the algebra
$\mathcal{L}_{n,\sigma}$. Hence, $\mathcal{L}_{n,\sigma}$ is the same for any
choice of $\sigma$, and one can just consider $\mathcal{L}_{n,q}$. For more
details on linearized polynomials we refer to [14, Chapter 3, Section 4].

Consider the non-degenerate symmetric bilinear form of $\mathbb{F}_{q^{n}}$
over $\mathbb{F}_{q}$ defined by

$\langle x,y\rangle=\mathrm{Tr}_{q^{n}/q}(xy),$

for every $x,y\in\mathbb{F}_{q^{n}}$ where
$\mathrm{Tr}_{q^{n}/q}(x)=x+\ldots+x^{q^{n-1}}$. The _adjoint_ $\hat{f}$ of
the $\sigma$-polynomial $\displaystyle
f=\sum_{i=0}^{n-1}a_{i}X^{\sigma^{i}}\in\mathcal{L}_{n,\sigma}$ with respect
to the bilinear form $\langle\cdot,\cdot\rangle$, i.e. the unique function
over $\mathbb{F}_{q^{n}}$ satisfying

$\mathrm{Tr}_{q^{n}/q}(yf(z))=\mathrm{Tr}_{q^{n}/q}(z\hat{f}(y))$

for every $y,z\in\mathbb{F}_{q^{n}}$, is given by

(3) $\hat{f}=\sum_{i=0}^{n-1}a_{i}^{\sigma^{n-i}}X^{\sigma^{n-i}}.$

In [1, Lemma 2.6] it has been proved that a $\sigma$-polynomial $f$ and its
adjoint always define the same linear set, and following the proof of [1,
Lemma 2.6] one also obtain that the weight distribution of $L_{f}$ coincides
with the weight distribution of $L_{\hat{f}}$; see also [18, Theorem 1]. More
precisely, the following holds (which we prove for sake of completeness in a
different way).

######  Lemma 2.2.

Let $f$ be a $\sigma$-polynomial in $\mathcal{L}_{n,\sigma}$. Then

(4)
$\\#\left\\{x\in\mathbb{F}_{q^{n}}^{*}:\frac{f(x)}{x}=b\right\\}=\\#\left\\{y\in\mathbb{F}_{q^{n}}^{*}:\frac{\hat{f}(y)}{y}=b\right\\}.$

In particular, $L_{f}=L_{\hat{f}}$ and $w_{L_{f}}(P)=w_{L_{\hat{f}}}(P)$ for
every point $P\in\mathrm{PG}(1,q^{n})$.

###### Proof.

Let $\chi(x)=\zeta_{p}^{\mathrm{Tr}_{q^{n}/p}(x)}$, where $p$ is the
characteristic of $\mathbb{F}_{q^{n}}$ and $\zeta_{p}$ is a $p$-th root of
unity in $\mathbb{C}$. Define $\chi_{a}(x)=\chi(ax)$ for
$a\in\mathbb{F}_{q^{n}}$. Then the character group of the additive group of
$\mathbb{F}_{q^{n}}$ is $\\{\chi_{a}:a\in\mathbb{F}_{q^{n}}\\}$. For
characters of groups, we refer to [11, Chapter 8].

By the definition of adjoints, for any $b\in\mathbb{F}_{q^{n}}$,

(5)
$\sum_{y\in\mathbb{F}_{q^{n}}}\sum_{x\in\mathbb{F}_{q^{n}}}\chi\left((f(x)-bx)y\right)=\sum_{x\in\mathbb{F}_{q^{n}}}\sum_{y\in\mathbb{F}_{q^{n}}}\chi\left((\hat{f}(y)-by)x\right).$

By the orthogonality of characters, the left-hand-side of (5) equals

$\\#\\{x\in\mathbb{F}_{q^{n}}:f(x)=bx\\}\cdot
q^{n}=\left(\\#\left\\{x\in\mathbb{F}_{q^{n}}^{*}:\frac{f(x)}{x}=b\right\\}+1\right)\cdot
q^{n}.$

The same result holds for the right-hand-side of (5). Therefore (4) is proved
and the last part of the assertion is a direct consequence. ∎

Given two arbitrary linearized polynomials $f$ and $g$, it is usually quite
difficult to determine whether $L_{f}=L_{g}$. However, we can obtain some
conditions involving the coefficients of both $f$ and $g$. Indeed, in [7,
Lemma 3.4] the following condition arises

$\sum_{x\in\mathbb{F}_{q^{n}}}\left(\frac{f(x)}{x}\right)^{e}=\sum_{x\in\mathbb{F}_{q^{n}}}\left(\frac{g(x)}{x}\right)^{e},$

for any positive integer $e$. For some special values of $e$, Csajbók, Marino
and Polverino [7] have provided some more explicit conditions.

######  Lemma 2.3.

[7, Lemma 3.6] Let $f=\sum_{i=0}^{n-1}\alpha_{i}X^{q^{i}}$ and
$g=\sum_{i=0}^{n-1}\beta_{i}X^{q^{i}}$ be two $q$-polynomials over
$\mathbb{F}_{q^{n}}$ such that $L_{f}=L_{g}$. Then

(6) $\alpha_{0}=\beta_{0}$

and

(7) $\alpha_{k}\alpha_{n-k}^{q^{k}}=\beta_{k}\beta_{n-k}^{q^{k}}$

for $k\in\\{1,2,\cdots,n-1\\}$, and

$\alpha_{1}\alpha_{k-1}^{q}\alpha_{n-k}^{q^{k}}+\alpha_{k}\alpha_{n-1}^{q}\alpha_{n-k+1}^{q^{k}}=\beta_{1}\beta_{k-1}^{q}\beta_{n-k}^{q^{k}}+\beta_{k}\beta_{n-1}^{q}\beta_{n-k+1}^{q^{k}},$

for $k\in\\{2,3,\cdots,n-1\\}$.

For some linearized polynomials with two terms, we can obtain a further
necessary condition.

######  Lemma 2.4.

Let $s$ and $n$ be two relatively prime positive integers. Let $\sigma\colon
x\in\mathbb{F}_{q^{n}}\mapsto x^{q^{s}}\in\mathbb{F}_{q^{n}}$. For any nonzero
elements $\theta,\delta,d\in\mathbb{F}_{q^{n}}$, let $f=X^{\sigma}+\theta
X^{\sigma^{n-1}}$ and $g=d(X^{\sigma}+\delta X^{\sigma^{n-1}})$ be two
$q$-polynomials over $\mathbb{F}_{q^{n}}$ such that $L_{f}=L_{g}$. Then

(8) $1+N_{q^{n}/q}(\theta)=N_{q^{n}/q}(d)(1+N_{q^{n}/q}(\delta)),$

when $n$ is odd, and

(9)
$1+N_{q^{n}/q}(\theta)+N_{q^{n}/q^{2}}(\theta)+N_{q^{n}/q^{2}}(\theta)^{\sigma}=N_{q^{n}/q}(d)(1+N_{q^{n}/q}(\delta)+N_{q^{n}/q^{2}}(\delta)+N_{q^{n}/q^{2}}(\delta)^{\sigma}),$

when $n$ is even.

###### Proof.

The $(1+q+q^{2}+\cdots+q^{n-1})$-th power of $f(X)/X$ modulo
$X^{\sigma^{n}}-X$ equals,

(10) $(X^{\sigma-1}+\theta
X^{\sigma^{n-1}-1})(X^{\sigma^{2}-\sigma}+\theta^{\sigma}X^{\sigma^{n}-\sigma})\cdots(X^{\sigma^{n}-\sigma^{n-1}}+\theta^{\sigma^{n-1}}X^{\sigma^{n-2}-\sigma^{n-1}}).$

As $\gcd(s,n)=1$, the coefficient of $X^{q^{n}-1}$ in (10) modulo
$X^{q^{n}}-X$ is the same as the coefficient of $X^{\sigma^{n}-1}$ modulo
$X^{\sigma^{n}}-X$.

To get $X^{\sigma^{n}-1}$ in (10), we only have to find all the possible ways
to choose exactly one term $Z_{i}$ from each set
$\\{X^{\sigma^{i+1}-\sigma^{i}},X^{\sigma^{i-1}-\sigma^{i}}\\}$ such that
$\prod_{i=0}^{n-1}Z_{i}\equiv X^{\sigma^{n}-1}\pmod{X^{\sigma^{n}}-X}$. To
determine it, we may consider the two terms $X^{\sigma^{i+1}-\sigma^{i}}$ and
$X^{\sigma^{i-1}-\sigma^{i}}$ in each
$(X^{\sigma^{i+1}-\sigma^{i}}+\theta^{\sigma}X^{\sigma^{i-1}-\sigma^{i}})$ as
two arrays of length $n$ defined recursively:

$\displaystyle\mathbf{a}_{0}$
$\displaystyle=(-1,1,0,\cdots,0)\in\mathbb{Z}^{n},$
$\displaystyle\mathbf{b}_{0}$
$\displaystyle=(-1,0,\cdots,0,1)\in\mathbb{Z}^{n},$

and for $i\in\\{1,2,\cdots,n-1\\}$, $\mathbf{a}_{i}$ (resp. $\mathbf{b}_{i}$)
is obtained from $\mathbf{a}_{i-1}$ (resp. $\mathbf{b}_{i-1}$) by applying a
shift on the right on its entries, that is

$\displaystyle\mathbf{a}_{i}$
$\displaystyle=(\overbrace{0,\cdots,0}^{i},-1,1,0,\cdots,0)\in\mathbb{Z}^{n},$
$\displaystyle\mathbf{b}_{i}$
$\displaystyle=(\overbrace{0,\cdots,0}^{i-1},1,-1,0,\cdots,0)\in\mathbb{Z}^{n}.$

In this way, $\mathbf{a}_{i}$ represents the exponents of $\sigma$ in the term
$X^{\sigma^{i+1}-\sigma^{i}}$ and, similarly, $\mathbf{b}_{i}$ represents the
exponents of $\sigma$ in the term $X^{\sigma^{i-1}-\sigma^{i}}$, for any
$i\in\\{0,\ldots,n-1\\}$. Since our aim is to find the coefficient of
$X^{\sigma^{n}-1}$, we have to find all the possible sum in which every addend
is exactly one element from $\mathbf{a}_{i}$ and $\mathbf{b}_{i}$ for each
$i\in\\{0,\cdots,n-1\\}$ and such that the sum equals to the zero vector
$\mathbf{0}$, which represents $X^{\sigma^{n}-1}$. For any $n$, it is easy to
see that

(11)
$\sum_{i=0}^{n-1}\mathbf{a}_{i}=\sum_{i=0}^{n-1}\mathbf{b}_{i}=\mathbf{0}.$

When $n$ is even, we also have

(12) $\sum_{2\mid i}\mathbf{a}_{i}+\sum_{2\nmid j}\mathbf{b}_{j}=\sum_{2\nmid
i}\mathbf{a}_{i}+\sum_{2\mid j}\mathbf{b}_{j}=\mathbf{0}.$

Claim: All the possible ways to get $\mathbf{0}$ are given in (11) and (12).

To prove the claim, we first have to notice that the two nonzero elements in
$\mathbf{a}_{i}$ or $\mathbf{b}_{i}$ are adjacent to each other. Consequently,
if $\mathbf{a}_{i}$ appears in the sum, to make the $(i+2)$-th position in the
array $0$, we have to add exactly one element from
$\\{\mathbf{a}_{i+1},\mathbf{b}_{i+1}\\}$. Moreover, if

$\mathbf{a}_{i}+\mathbf{a}_{i+1}=(\overbrace{0,\cdots,0}^{i},-1,0,1,0,\cdots,)$

is contained in the sum, then it is easy to see that the $(i-1)$-th term of
the sum has to be $\mathbf{a}_{i-1}$. The same result also holds for
$\mathbf{b}_{i}+\mathbf{b}_{i+1}$. Whereas, if
$\mathbf{a}_{i}+\mathbf{b}_{i+1}$ is contained in the sum, since
$\mathbf{a}_{i}+\mathbf{b}_{i+1}=\mathbf{0}$, then we necessarily obtain the
case (12). As a consequence, we complete the proof of the claim.

By Claim, the coefficient of $X^{q^{n}-1}$ in (10) is

$c_{q^{n}-1}=\begin{cases}1+N_{q^{n}/q}(\theta),&\text{ if }2\nmid n;\\\
1+N_{q^{n}/q}(\theta)+N_{q^{n}/q^{2}}(\theta)+N_{q^{n}/q^{2}}(\theta)^{\sigma},&\text{
if }2\mid n.\end{cases}$

By the well known equation

$\sum_{x\in\mathbb{F}_{q^{n}}}x^{e}=\begin{cases}-1,&q^{n}-1\mid e;\\\
0,&\text{otherwise},\end{cases}$

we have

$\sum_{x\in\mathbb{F}_{q^{n}}}(x^{q^{s}-1}+\theta
x^{q^{n-s}-1})^{1+q+\cdots+q^{n-1}}=c_{q^{n}-1}\sum_{x\in\mathbb{F}_{q^{n}}}x^{q^{n}-1}=-c_{q^{n}-1}.$

As $L_{g}=L_{f}$, by comparing the coefficients of $X^{q^{n}-1}$ of
$(f(X)/X)^{1+q+\cdots+q^{n-1}}$ and $(g(X)/X)^{1+q+\cdots+q^{n-1}}$ modulo
$X^{q^{n}}-X$, we get (8) and (9). ∎

##  3\. Equivalence of the Lunardon-Polverino linear sets and their
automorphism group

In this section, we completely determine the equivalence between different
members of the Lunardon-Polverino maximum scattered linear sets in a
projective line as well as their automorphism groups. The main results of this
section are the following.

######  Theorem 3.1.

Let $n,s,t\in\mathbb{Z}$ such that $\gcd(s,n)=\gcd(t,n)=1$, $n\geq 3$ and
$1\leq s,t<n/2$. For any nonzero elements $\theta,\delta\in\mathbb{F}_{q^{n}}$
satisfying $N_{q^{n}/q}(\theta),N_{q^{n}/q}(\delta)\neq 1$, let
$f=X^{q^{s}}+\theta X^{q^{s(n-1)}}$ and $g=X^{q^{t}}+\delta X^{q^{t(n-1)}}$.
The linear sets $L_{f}$ and $L_{g}$ are equivalent if and only if $s=t$ and
one of the following collections of conditions is satisfied:

  * (a)

$2\nmid n$,
$N_{q^{n}/q}(\theta)\in\\{N_{q^{n}/q}(\delta^{\tau}),N_{q^{n}/q}(1/\delta^{\tau})\\}$
for some $\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$

  * (b)

$2\mid n$,
$N_{q^{n}/q^{2}}(\theta)\in\\{N_{q^{n}/q^{2}}(\delta^{\tau}),N_{q^{n}/q^{2}}(1/\delta^{\tau})\\}$
for some $\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$.

######  Theorem 3.2.

Let $n,s\in\mathbb{Z}$ such that $\gcd(s,n)=1$, $n\geq 3$ and $1\leq s<n/2$.
For any nonzero element $\theta\in\mathbb{F}_{q^{n}}$ satisfying
$N_{q^{n}/q}(\theta)\neq 1$, let $f=X^{q^{s}}+\theta X^{q^{s(n-1)}}$. Then the
$\mathrm{P\Gamma L}$-automorphism group of $L_{f}$ is

$\mathrm{Aut}(L_{f})=\begin{cases}\mathcal{D},&n>4;\\\
\mathcal{D}\cup\mathcal{C},&n=4,\end{cases}$

where

(13) $\mathcal{D}:=\left\\{\left(\begin{pmatrix}1&0\\\
0&d\end{pmatrix},\tau\right)\in\mathrm{P\Gamma
L}(2,q^{n}):d^{\sigma+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{\sigma}\right\\}$

and

(14) $\mathcal{C}:=\left\\{\left(\begin{pmatrix}0&1\\\
c&0\end{pmatrix},\tau\right)\in\mathrm{P\Gamma
L}(2,q^{4}):\left(\frac{c}{\theta^{\tau q}-\theta^{-\tau
q^{3}}}\right)^{q+1}=\left(-\theta^{\tau q^{3}+1}\right)^{q}\right\\}.$

In particular, its size is

(15) $\\#\mathrm{Aut}(L_{f})=\begin{cases}N_{\tau}(\theta),&2\nmid n,2\mid
q;\\\ 2N_{\tau}(\theta),&2\nmid n,2\nmid q;\\\ (q+1)N_{\tau}(\theta),&2\mid
n,n>4;\\\ 2(q+1)N_{\tau}(\theta),&n=4,\end{cases}$

where

$N_{\tau}(\theta):=\\#\left\\{\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}}):N_{q^{n}/q}(\theta)=N_{q^{n}/q}(\theta^{\tau})\text{
or }1/N_{q^{n}/q}(\theta^{\tau})\right\\}$

for odd $n$ and

$N_{\tau}(\theta):=\\#\left\\{\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}}):N_{q^{n}/q^{2}}(\theta)=\\{N_{q^{n}/q^{2}}(\theta^{\tau})\text{
or }1/N_{q^{n}/q^{2}}(\theta^{\tau})\right\\}$

for even $n$.

######  Remark 3.3.

Let $n,s\in\mathbb{Z}$ such that $\gcd(s,n)=1$, $s>n/2$. It is easy to see
that the linear set $L_{f}$ defined by $f=X^{q^{s}}+\theta X^{q^{s(n-1)}}$ is
equivalent to $L_{g}$ where $g=X^{q^{n-s}}+\delta X^{q^{(n-s)(n-1)}}$, with
$\delta=\theta^{-1}$. So, the assumptions in Theorem 3.1 on $s$ and $t$ are
not restrictive.

In order to get Theorems 3.1 and 3.2, we need to prove a series of lemmas.

######  Lemma 3.4.

Let $s$ and $n$ be two relatively prime positive integers, and $1\leq s<n/2$.
Let $\sigma\colon x\in\mathbb{F}_{q^{n}}\mapsto
x^{q^{s}}\in\mathbb{F}_{q^{n}}$. For any nonzero elements
$\theta,\delta\in\mathbb{F}_{q^{n}}$ satisfying
$N_{q^{n}/q}(\theta),N_{q^{n}/q}(\delta)\neq 1$, let $f=X^{\sigma}+\theta
X^{\sigma^{n-1}}$ and $g=X^{\sigma}+\delta X^{\sigma^{n-1}}$. There exists
$d\in\mathbb{F}_{q^{n}}$ such that the linear sets $L_{f}=L_{dg}$ if there
exists $\alpha\in\mathbb{F}_{q^{n}}$ such that

(16) $\alpha^{\sigma^{2}-1}=\left(\frac{\theta}{\delta}\right)^{\sigma}.$

###### Proof.

Replace $X$ by $\alpha X$ in $f(X)/X$. Let $d=\alpha^{\sigma-1}$. Then we
obtain

$\frac{f(\alpha X)}{\alpha
X}=\alpha^{\sigma-1}X^{\sigma-1}+\theta\alpha^{\sigma^{n-1}-1}X^{\sigma^{n-1}-1}=d\frac{g(X)}{X},$

since
$(d\delta)^{\sigma}=\alpha^{\sigma^{2}-\sigma}\delta^{\sigma}=\theta^{\sigma}\alpha^{1-\sigma}$,
that is $d\delta=\theta\alpha^{\sigma^{n-1}-1}$. Therefore, $L_{f}=L_{dg}$. ∎

######  Remark 3.5.

The sufficient condition given in Lemma 3.4 is actually the necessary and
sufficient condition for the existences of permutation $q$-polynomials $L_{1}$
and $L_{2}$ such that $\mathcal{C}_{g}=\\{L_{1}\circ h\circ
L_{2}(X):h\in\mathcal{C}_{f}\\}$, where

$\mathcal{C}_{f}:=\\{aX+bf(X):a,b\in\mathbb{F}_{q^{n}}\\}.$

See [17, Definition 4.1 and Theorem 4.4] for more details. By Hilbert’s
theorem 90, the condition given in (16) is equivalent to

(17) $\begin{cases}N_{q^{n}/q}(\theta)=N_{q^{n}/q}(\delta),&2\nmid n;\\\
N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q^{2}}(\delta),&2\mid n.\end{cases}$

On the other hand, we can get some necessary condition for $L_{f}=L_{dg}$ for
some $d\in\mathbb{F}_{q^{n}}$ which is slightly different from (17).

######  Lemma 3.6.

Let $s$ and $n$ be two relatively prime positive integers, and $1\leq s<n/2$.
Let $\sigma\colon x\in\mathbb{F}_{q^{n}}\mapsto
x^{q^{s}}\in\mathbb{F}_{q^{n}}$. For any nonzero elements
$\theta,\delta\in\mathbb{F}_{q^{n}}$ satisfying
$N_{q^{n}/q}(\theta),N_{q^{n}/q}(\delta)\neq 1$, let $f=X^{\sigma}+\theta
X^{\sigma^{n-1}}$ and $g=X^{\sigma}+\delta X^{\sigma^{n-1}}$. There exists
$d\in\mathbb{F}_{q^{n}}$ such that the linear sets $L_{f}=L_{dg}$ if and only
if

(18) $N_{q^{n}/q}(\theta)\in\\{N_{q^{n}/q}(\delta),N_{q^{n}/q}(1/\delta)\\},$

when $n$ is odd, and

(19)
$N_{q^{n}/q^{2}}(\theta)\in\\{N_{q^{n}/q^{2}}(\delta),N_{q^{n}/q^{2}}(1/\delta)^{\sigma}\\},$

when $n$ is even.

###### Proof.

First we show that (18) and (19) are sufficient. As we have proved in Remark
3.5 that (17) is equivalent to (16), if
$N_{q^{n}/q}(\theta)=N_{q^{n}/q}(\delta)$ for $2\nmid n$ and
$N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q^{2}}(\delta)$ for $2\mid n$, then there
exists $d\in\mathbb{F}_{q^{n}}$ such that $L_{f}=L_{dg}$.

By (3), the adjoint of $f$ is
$\hat{f}=\theta^{\sigma}X^{\sigma}+X^{\sigma^{n-1}}$. By Lemma 2.2,
$L_{f}=L_{\hat{f}}$ and again we may apply Remark 3.5 and Lemma 3.4, obtaining
the existence of $d^{\prime}\in\mathbb{F}_{q^{n}}$ such that

$\left\\{x^{\sigma-1}+\frac{1}{\theta^{\sigma}}x^{\sigma^{n-1}-1}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{d^{\prime}\left(x^{\sigma-1}+{\delta}x^{\sigma^{n-1}-1}\right):x\in\mathbb{F}_{q^{n}}^{*}\right\\},$

provided that $N_{q^{n}/q}(\theta)=N_{q^{n}/q}(1/\delta)$ for $n$ odd and
$N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q^{2}}(1/\delta)^{\sigma}$ for $n$ even.
Therefore, we finish the proof of the sufficiency.

Next we prove the necessity of (18) and (19). By (7) for $k=s$, we get

$\theta^{\sigma}=d^{\sigma+1}\delta^{\sigma},$

i.e.

(20) $d^{\sigma+1}=\left(\frac{\theta}{\delta}\right)^{\sigma}.$

When $2\nmid n$, (20) implies

$N_{q^{n}/q}(d)^{2}=N_{q^{n}/q}\left(\frac{\theta}{\delta}\right).$

Plugging it into (8), we have

$1+N_{q^{n}/q}(\delta)N_{q^{n}/q}(d)^{2}=N_{q^{n}/q}(d)\left(1+N_{q^{n}/q}(\delta)\right),$

which is

(21)
$N_{q^{n}/q}(d)^{2}-\left(1+\frac{1}{N_{q^{n}/q}(\delta)}\right)N_{q^{n}/q}(d)+\frac{1}{N_{q^{n}/q}(\delta)}=0,$

that is $N_{q^{n}/q}(d)$ is a root of

$X^{2}-\left(1+\frac{1}{N_{q^{n}/q}(\delta)}\right)X+\frac{1}{N_{q^{n}/q}(\delta)}.$

It follows that $N_{q^{n}/q}(d)=1$ or $N_{q^{n}/q}(d)=1/N_{q^{n}/q}(\delta)$
which implies (18).

When $2\mid n$, (20) implies

(22)
$N_{q^{n}/q}(d)=N_{q^{n}/q}(d)^{\sigma}=N_{q^{n}/q^{2}}\left(\frac{\theta}{\delta}\right).$

Hence $N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q}(d)N_{q^{n}/q^{2}}(\delta)$ and
$N_{q^{n}/q}(\theta)=N_{q^{n}/q}(d)^{2}N_{q^{n}/q}(\delta)$. Plugging it into
(9), we have

$\displaystyle
1+N_{q^{n}/q}(d)^{2}N_{q^{n}/q}(\delta)+N_{q^{n}/q}(d)N_{q^{n}/q^{2}}(\delta)+N_{q^{n}/q}(d)N_{q^{n}/q^{2}}(\delta)^{\sigma}$
$\displaystyle=$ $\displaystyle
N_{q^{n}/q}(d)(1+N_{q^{n}/q}(\delta)+N_{q^{n}/q^{2}}(\delta)+N_{q^{n}/q^{2}}(\delta)^{\sigma}),$

which is (21) again. It follows that $N_{q^{n}/q}(d)=1$ or
$N_{q^{n}/q}(d)=1/N_{q^{n}/q}(\delta)$. Together with (22), we get
$N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q^{2}}(\delta)$ or
$N_{q^{n}/q^{2}}(\theta)=N_{q^{n}/q^{2}}(1/\delta)^{\sigma}$. ∎

######  Remark 3.7.

In the proof of Lemma 3.6, when (18) or (19) is satisfied, we see that

(23)
$\\{d\in\mathbb{F}_{q^{n}}:L_{f}=L_{dg}\\}=\left\\{d\in\mathbb{F}_{q^{n}}:d^{\sigma+1}=\left(\frac{\theta}{\delta}\right)^{\sigma}\right\\}.$

Consequently, by Lemma 2.1, there are

(24) $\gcd(q^{s}+1,q^{n}-1)=\gcd(q+1,q^{n}-1)=\begin{cases}1,&2\nmid n,2\mid
q;\\\ 2,&2\nmid n,2\nmid q;\\\ q+1,&2\mid n,\end{cases}$

possible choices of $d$ such that $L_{f}=L_{dg}$.

######  Lemma 3.8.

Let $n,s,t\in\mathbb{Z}$ such that $\gcd(s,n)=\gcd(t,n)=1$, and $1\leq
s,t<n/2$. For any integer $n>4$ and prime power $q$, define
$f=X^{q^{s}}+\theta X^{q^{n-s}}$ and $g=X^{q^{t}}+\delta X^{q^{n-t}}$ with
$N_{q^{n}/q}(\theta),N_{q^{n}/q}(\delta)\notin\\{0,1\\}$. If the map $x\mapsto
f(x)$ is bijective on $\mathbb{F}_{q^{n}}$, then there is no nonzero constant
$c$ such that

(25)
$\left\\{\frac{cx}{f(x)}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{\frac{g(x)}{x}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}.$

###### Proof.

Suppose that the inverse map of $f(x)$ is
$x\mapsto\sum_{i=0}^{n-1}r_{i}x^{q^{i}}$ with $r_{i}\in\mathbb{F}_{q^{n}}$.
Consequently,

$\left(\sum_{i=0}^{n-1}r_{i}X^{q^{i}}\right)^{q^{s}}+\theta\left(\sum_{i=0}^{n-1}r_{i}X^{q^{i}}\right)^{q^{n-s}}\equiv
X\pmod{X^{q^{n}}-X},$

which means

(26) $\sum_{i=0}^{n-1}\left(r_{i-s}^{q^{s}}+\theta
r_{i+s}^{q^{n-s}}\right)X^{q^{i}}\equiv X\pmod{X^{q^{n}}-X},$

where the subscript of each $r_{i}$ is computed modulo $n$. Note that

(27)
$\left\\{\frac{cx}{f(x)}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{\frac{cf^{-1}(x)}{x}:x\in\mathbb{F}_{q^{n}}^{*}\right\\},$

since $f$ is invertible. Comparing the coefficients of each term in (26), we
get

(28) $r_{n-s}^{q^{s}}+\theta r_{s}^{q^{n-s}}=1,$

and

(29) $r_{i-s}^{q^{s}}=-\theta r_{i+s}^{q^{n-s}},i\in\\{1,2,3,...n-1\\}.$

Suppose to the contrary that there exists a nonzero constant $c$ such that
(25) is satisfied. Applying Lemma 2.3 on (25), taken into account (27), we see

$r_{0}=0.$

For $i\in\\{1,2,\cdots,n-1\\}$, if $r_{i-s}=0$, then by (29) $r_{i+s}=0$. Thus
$r_{2s}=0$, because $r_{s-s}=r_{0}=0$. By induction,

(30) $r_{2\ell s}=0\text{ for }\ell=0,1,2,\cdots.$

If $n$ is odd, then (30) implies $r_{2\cdot\frac{n-1}{2}s}=r_{n-s}=0$ and
$r_{2\cdot\frac{n+1}{2}s}=r_{s}=0$ which contradicts (28).

If $n$ is even, then (30) implies $r_{2i}=0$ for all $i=0,1,\cdots,n/2-1$. By
(7),

$r_{i}r_{n-i}^{q^{i}}=0,$

for $i\neq t,n-t$, which are equivalent to $r_{i}=0$ or $r_{n-i}=0$. As $n>4$,
there always exists an odd integer $i_{0}$ satisfying $i_{0}\neq t,n-t$ and
$r_{i_{0}}=0$. By induction using (29), $r_{2\ell s+i_{0}}=0$ for
$\ell=0,1,\cdots$ which implies $r_{s}=r_{n-s}=0$ because $s$ and $n-s$ are
both odd. This again contradicts (28). ∎

For $n=4$, Lemma 3.8 does not hold any more.

######  Lemma 3.9.

For prime power $q$, define $f=X^{q}+\theta X^{q^{3}}$ and $g=X^{q}+\delta
X^{q^{3}}$ over $\mathbb{F}_{q^{4}}$ with
$N_{q^{4}/q}(\theta),N_{q^{4}/q}(\delta)\notin\\{0,1\\}$. There exists $c$
such that

(31)
$\left\\{\frac{cx}{f(x)}:x\in\mathbb{F}_{q^{4}}^{*}\right\\}=\left\\{\frac{g(x)}{x}:x\in\mathbb{F}_{q^{4}}^{*}\right\\}$

if and only if
$\theta^{q^{2}+1}\in\\{\delta^{q^{2}+1},\delta^{-(q^{3}+q)}\\}$.

When one of the above conditions is satisfied, the number of
$c\in\mathbb{F}_{q^{4}}$ such that (31) holds is $q+1$.

###### Proof.

Let $h=X^{q}-\theta^{-q^{3}}X^{q^{3}}$. Then

$h(f(X))=(\theta^{q}-\theta^{-q^{3}})X.$

Thus $x\mapsto\frac{h(x)}{\theta^{q}-\theta^{-q^{3}}}$ is the inverse map of
$f(x)$, and (31) equals

$\left\\{d\cdot\frac{h(x)}{x}:x\in\mathbb{F}_{q^{4}}^{*}\right\\}=\left\\{\frac{g(x)}{x}:x\in\mathbb{F}_{q^{4}}^{*}\right\\},$

where $d=\frac{c}{\theta^{q}-\theta^{-q^{3}}}$.

By Lemma 3.6 for $n=4$, there exists such a $d\in\mathbb{F}_{q^{4}}$ if and
only if
$N_{q^{4}/q^{2}}(\theta^{-q^{3}})=\theta^{-(q^{3}+q)}=\delta^{q^{2}+1}$ or
$\delta^{-(q^{3}+q)}$, which is equivalent to
$\theta^{q^{2}+1}=\delta^{q^{2}+1}$ or $\delta^{-(q^{3}+q)}$.

The last part of the statement of Lemma 3.9 follows from Remark 3.7. ∎

######  Remark 3.10.

Comparing the conditions in Lemma 3.8 and Lemma 3.9, we notice that we have
additionally assumed that $f$ is bijective on $\mathbb{F}_{q^{n}}$ in Lemma
3.8. In fact, when $n$ is even, $f$ is always bijective. Consider
$f=X^{q^{s}}+\theta X^{q^{n-s}}\in\mathcal{L}_{n,q}$ with $\gcd(n,s)=1$ and
$N_{q^{n}/q}(\theta)\notin\\{0,1\\}$. Suppose that $f$ is not bijective, that
is there exists $x\in\mathbb{F}_{q^{n}}^{*}$ such that

$\theta=-x^{q^{s}-q^{n-s}}.$

This happens if and only in $N_{q^{n}/q^{e}}(\theta)=N_{q^{n}/q^{e}}(-1)$,
where $e=\gcd(n,n-2s)=\gcd(n,2)$. If $n$ is even then
$N_{q^{n}/q}(\theta)=N_{q^{n}/q}(-1)=1$, which is a contradiction to the fact
that $N_{q^{n}/q}(\theta)\notin\\{0,1\\}$. If $n$ is odd then $e=1$ and
$N_{q^{n}/q}(\theta)=-1$ and in this case $f$ has $q$ roots over
$\mathbb{F}_{q^{n}}$. Therefore, $f$ is bijective if and only if $n$ is even
or $n$ is odd and $N_{q^{n}/q^{e}}(\theta)\neq-1$.

Now we are ready to prove Theorem 3.1.

###### Proof of Theorem 3.1.

Assume that $L_{f}$ and $L_{g}$ are equivalent which implies the existence of
$M=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ over $\mathbb{F}_{q^{n}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$ such that

(32) $\left\\{x^{q^{s}-1}+\theta
x^{q^{n-s}-1}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{\frac{cx+d\overline{g}(x)}{ax+b\overline{g}(x)}:x\in\mathbb{F}_{q^{n}}^{*}\right\\},$

where $\overline{g}(x)=x^{q^{t}}+\delta^{\tau}x^{q^{n-t}}$.

Depending on the value of $b$, we separate the rest part into two cases.

Case 1: $b=0$. As $M$ is viewed as an element in $\mathrm{PGL}(2,q^{n})$, we
can assume that $a=1$ and (32) becomes

(33) $\left\\{x^{q^{s}-1}+\theta
x^{q^{n-s}-1}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{c+d\frac{\overline{g}(x)}{x}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}.$

By (6) in Lemma 2.3, we derive $c=0$. By (7) for $k=s$, we can get $s=t$;
otherwise $\theta^{q^{t}}=0$ which contradicts the assumption that
$N_{q^{n}/q}(\theta)\neq 0$. Now (33) is exactly $L_{f}=L_{d\overline{g}}$. By
Lemma 3.6, we derive (a) and (b) of Theorem 3.1 as necessary conditions.

Case 2: $b\neq 0$. Without loss of generality, we assume that $b=1$. We are
going to prove the following claim:

Claim: $d=0$.

Suppose to the contrary that $d\neq 0$. Now

$\frac{cx+d\overline{g}(x)}{ax+b\overline{g}(x)}=d+\frac{(c-ad)x}{ax+\overline{g}(x)},$

for $x\in\mathbb{F}_{q^{n}}$. By definition of the equivalence of linear sets,
the map $x\mapsto ax+\overline{g}(x)$ is invertible (otherwise
$\langle(0,1)\rangle_{\mathbb{F}_{q^{n}}}$ would be in $L_{f}$) and $c-ad\neq
0$. Let $h(x)=\sum_{i=0}^{n-1}r_{i}x^{q^{i}}$ denote the inverse map of
$x\mapsto\frac{ax+\overline{g}(x)}{c-ad}$. Setting
$y=\frac{ax+\overline{g}(x)}{c-ad}$, we get

$d+\frac{(c-ad)x}{ax+\overline{g}(x)}=d+\frac{h(y)}{y},$

for $x\in\mathbb{F}_{q^{n}}$. By the definition of $h$ and by comparing the
coefficients of $ah(Y)+\overline{g}(h(Y))\equiv(c-ad)Y\pmod{Y^{q^{n}}-Y}$, we
obtain

(34) $ar_{0}+r_{n-t}^{q^{t}}+\delta^{\tau}r_{t}^{q^{n-t}}=c-ad$

and

(35) $ar_{i}+r_{i-t}^{q^{t}}+\delta^{\tau}r_{i+t}^{q^{n-t}}=0,$

for $i\in\\{1,2,\cdots,n-1\\}$.

Now (32) becomes

(36) $\left\\{x^{q^{s}-1}+\theta
x^{q^{n-s}-1}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=\left\\{d+\sum_{i=0}^{n-1}r_{i}y^{q^{i}-1}:y\in\mathbb{F}_{q^{n}}^{*}\right\\}.$

By (6) of Lemma 2.3,

(37) $r_{0}=-d.$

Moreover, by (7) of Lemma 2.3,

(38) $r_{k}r_{n-k}^{q^{k}}=\begin{cases}\theta^{q^{s}},&k=s;\\\
\theta,&k=n-s;\\\ 0,&k\neq 0,s,n-s.\end{cases}$

Depending on the value of $t$, we consider 3 cases. In each case, we will get
a contradiction which means the assumption $d\neq 0$ cannot hold.

Case 2.1: $s\neq\pm t,\pm 2t,\pm 3t\pmod{n}$. As $\gcd(n,s)=\gcd(n,t)=1$, $n$
must be larger than $4$.

By (38),

$r_{2t}r_{n-2t}^{q^{2t}}=0,$

which means at least one of $r_{2t}$ and $r_{n-2t}$ equals $0$. Without loss
of generality, we assume that $r_{n-2t}=0$. By (35) with $i=n-t$, that is

$ar_{n-t}+r_{n-2t}^{q^{t}}+\delta^{\tau}r_{0}^{q^{n-t}}=0,$

and $r_{0}=-d\neq 0$, we get $a\neq 0$ and $r_{n-t}\neq 0$. Together with (38)
for $k=t$, we derive that

$r_{t}=0.$

Plugging it into (35) with $i=t$, that is

$ar_{t}+r_{0}^{q^{t}}+\delta^{\tau}r_{2t}^{q^{n-t}}=0,$

we obtain $r_{2t}\neq 0$. Furthermore, as $r_{2t}\neq 0$ and $r_{t}=0$, (35)
with $i=2t$, that is

$ar_{2t}+r_{t}^{q^{t}}+\delta^{\tau}r_{3t}^{q^{n-t}}=0,$

implies $r_{3t}\neq 0$.

On the other hand, as $r_{n-t}\neq 0$ and $r_{n-2t}=0$, (35) with $i=n-2t$,
i.e.,

$ar_{n-2t}+r_{n-3t}^{q^{t}}+\delta^{\tau}r_{n-t}^{q^{n-t}}=0,$

implies $r_{n-3t}\neq 0$. Thus $r_{3t}r_{n-3t}\neq 0$. However, this
contradicts (38).

Case 2.2: $s\equiv\pm 2t\pmod{n}$. By Remark 3.3, we only have to consider the
case $s\equiv 2t\pmod{n}$. Since $t\neq\pm s\pmod{n}$, (38) with $k=t$ implies
that at least one of $r_{t}$ and $r_{n-t}$ is $0$. Without loss of generality,
we assume that $r_{t}=0$. Moreover, by (34) and $r_{0}=-d$,

(39) $r_{n-t}=c^{q^{n-t}}.$

By (35) with $i=t$, we get

$r_{0}^{q^{t}}+\delta^{\tau}r_{s}^{q^{n-t}}=0,$

which implies

(40) $r_{s}=\frac{d^{q^{2t}}}{\delta^{\tau q^{t}}}.$

By (38) with $k=n-s$, we have

(41) $r_{n-s}=\frac{\theta}{r_{s}^{q^{n-s}}}\neq 0.$

Plugging (39), (40) and (41) into the $q^{t}$-th power of

$ar_{n-t}+r_{n-2t}^{q^{t}}+\delta^{\tau}r_{0}^{q^{n-t}}=0,$

we have

(42) $a^{q^{t}}c+\delta^{\tau
q^{t}}\left(\frac{\theta}{d}\right)^{q^{2t}}-d\delta^{\tau q^{t}}=0.$

To show that (42) actually leads to contradiction, we are going to derive an
extra condition on the value of $c$ and $a$.

First, we can simply get $a\neq 0$: if $a=0$, then by taking $i=3t$ and
$i=-3t$ in (35), we obtain

$\displaystyle r_{2t}^{q^{t}}+\delta^{\tau}r_{4t}^{q^{n-t}}$
$\displaystyle=0,$ $\displaystyle
r_{n-4t}^{q^{t}}+\delta^{\tau}r_{n-2t}^{q^{n-t}}$ $\displaystyle=0,$

which implies $r_{4t}\neq 0$ and $r_{n-4t}\neq 0$. However, by formula (40),
(41) and the assumption $d\neq 0$, it is a contradiction with (38) for $k=4t$.

As $r_{t}=0$, (35) with $i=2t=s$ becomes

(43) $ar_{2t}+\delta^{\tau}r_{3t}^{q^{n-t}}=0.$

By $a\neq 0$, (40) and (43), $r_{3t}\neq 0$. By (38) with $k=3t$,
$r_{n-3t}=0$. Hence, (35) with $i=n-2t$ becomes

$ar_{n-2t}+\delta^{\tau}r_{n-t}^{q^{n-t}}=0.$

Together with (41) and (40), we obtain

$r_{n-t}^{q^{t}}=-\left(\frac{a}{\delta^{\tau}}\right)^{q^{2t}}\frac{\theta^{q^{2t}}}{r_{s}}=-\left(\frac{a}{\delta^{\tau}}\right)^{q^{2t}}\frac{\theta^{q^{2t}}}{d^{q^{2t}}}\delta^{\tau
q^{t}}.$

By (39), we get

(44) $c=-\left(\frac{a\theta}{d}\right)^{q^{2t}}\delta^{\tau(q^{t}-q^{2t})}.$

Setting $i=3t$ and $n-3t$ in (35), respectively, we get

$ar_{3t}+r_{2t}^{q^{t}}+\delta^{\tau}r_{4t}^{q^{n-t}}=0,$

and

$ar_{n-3t}+r_{n-4t}^{q^{t}}+\delta^{\tau}r_{n-2t}^{q^{n-t}}=0.$

Recall that $r_{n-3t}=0$ and $r_{n-s}\neq 0$. Consequently, the second
equation above implies that $r_{n-4t}\neq 0$. By (38), $r_{4t}=0$ and the
first equation above implies

$ar_{3t}=-r_{2t}^{q^{t}}.$

Together with (43), we have

(45) $a^{q^{t}+1}=\delta^{\tau q^{t}}.$

Plugging (44) and (45) into the left-hand side of (42),

$\displaystyle a^{q^{t}}c+\delta^{\tau
q^{t}}\left(\frac{\theta}{d}\right)^{q^{2t}}-d\delta^{\tau q^{t}}$
$\displaystyle=$
$\displaystyle-\frac{(a^{q^{t}+1})^{q^{t}}}{d^{q^{2t}}}\left(\frac{\theta}{\delta^{\tau}}\right)^{q^{2t}}\delta^{\tau
q^{t}}+\delta^{\tau q^{t}}\left(\frac{\theta}{d}\right)^{q^{2t}}-d\delta^{\tau
q^{t}}$ $\displaystyle=$ $\displaystyle-\frac{\delta^{\tau
q^{2t}}}{d^{q^{2t}}}\left(\frac{\theta}{\delta^{\tau}}\right)^{q^{2t}}\delta^{\tau
q^{t}}+\delta^{\tau q^{t}}\left(\frac{\theta}{d}\right)^{q^{2t}}-d\delta^{\tau
q^{t}}$ $\displaystyle=$ $\displaystyle-d\delta^{\tau q^{t}}.$

Therefore, $-d\delta^{\tau q^{t}}=0$. However, this is a contradiction with
the assumption $d\neq 0$ and $\delta\neq 0$.

Case 2.3: $s\equiv\pm 3t\pmod{n}$. Now we assume that $n>4$; otherwise $n=4$
and $s\equiv\pm t\pmod{n}$ which will be investigated later in the next case.
By Remark 3.3, we only need to consider the case $s\equiv 3t\pmod{n}$. Our
strategy is very similar to Case 2.2. As $t\neq\pm s\pmod{n}$, (38) with $k=t$
implies that at least one of $r_{t}$ and $r_{n-t}$ is $0$. Without loss of
generality, we assume that $r_{t}=0$. Moreover, by (34) and $r_{0}=-d\neq 0$,
we also have

(46) $r_{n-t}=c^{q^{n-t}}.$

By (35) with $i=t$, we derive

$r_{2t}=\frac{d^{q^{2t}}}{\delta^{\tau q^{t}}}.$

Together with (38), we get $r_{n-2t}=0$. Plugging it into (35) with $i=n-t$,
we have $a\neq 0$ and

$r_{n-t}=\frac{\delta^{\tau}d^{q^{n-t}}}{a}.$

From the above equation and (46), we derive

(47) $c=\frac{\delta^{\tau q^{t}}d}{a^{q^{t}}}.$

Plugging the value of $r_{2t}$, $r_{t}$ and $r_{n-2t}$, $r_{n-t}$ into (35)
for $i=2t$ and $i-n-2t$, respectively, we get

(48) $r_{3t}=-\frac{\delta^{\tau(q^{t}+q^{2t})}}{a^{q^{t}}d^{q^{3t}}},$

and

$r_{n-3t}=-\frac{\delta^{\tau(q^{n-t}+q^{n-2t})}}{a^{q^{n}-2t}}\cdot
d^{q^{n-3t}}.$

Another round of calculation with (35) for $i=3t$ and $i=n-3t$ leads to

(49) $r_{n-4t}=-(ar_{n-3t})^{q^{n-t}}\neq 0$

which by (38) implies $r_{4t}=0$, and

$r_{3t}=-\frac{d^{q^{3t}}}{a\delta^{\tau q^{2t}}}.$

The same computation with (35) for $i=4t$ and $i=n-4t$ provides us

$r_{5t}\neq 0,~{}~{}r_{n-5t}=0,$

and

$r_{n-4t}=-\frac{\delta^{\tau}}{a}r_{n-3t}^{q^{n-t}}.$

Together with (49), we get

$a^{q^{t}+1}=\delta^{\tau q^{t}}.$

However, if we plug it into (47), we obtain

$c=ad$

which contradicts the assumption that the matrix $M=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}$ with $b=1$ is invertible.

Case 2.4: $t\equiv\pm s\pmod{n}$. Again we only look at the case $t\equiv
s\pmod{n}$. By (38), $r_{t}$ and $r_{n-t}$ are both nonzero, and

$r_{2t}r_{n-2t}^{q^{2t}}=0.$

Without loss of generality, we assume that $r_{2t}=0$. By (35) with $i=2t$,
that is

$ar_{2t}+r_{t}^{q^{t}}+\delta^{\tau}r_{3t}^{q^{n-t}}=0,$

$r_{3t}=-\frac{r_{t}^{q^{2t}}}{\delta^{\tau q^{t}}}\neq 0$ whence $r_{n-3t}=0$
by (38). Consequently, (35) with $i=n-2t$, that is

$ar_{n-2t}+r_{n-3t}^{q^{t}}+\delta^{\tau}r_{n-t}^{q^{n-t}}=0,$

implies

(50) $ar_{n-2t}=-\delta^{\tau}\cdot r_{n-t}^{q^{n-t}}.$

Now let us look at (35) with $i=3t$ and $n-3t$,

$\displaystyle ar_{3t}+r_{2t}^{q^{t}}+\delta^{\tau}r_{4t}^{q^{n-t}}$
$\displaystyle=0,$ $\displaystyle
ar_{n-3t}+r_{n-4t}^{q^{t}}+\delta^{\tau}r_{n-2t}^{q^{n-t}}$ $\displaystyle=0.$

As $r_{2t}$ is assumed to be $0$ and $r_{3t}\neq 0$, the first equation above
implies $r_{4t}\neq 0$. Consequently, (38) means $r_{n-4t}=0$. Together with
$r_{n-3t}=0$ and the second equation above, we get $r_{n-2t}=0$. By (50) and
$\delta\neq 0$, we obtain

$r_{n-t}=0.$

However, this contradicts (38) with $k=t=s$ which implies that $r_{t}$ and
$r_{n-t}$ are both nonzero.

Therefore, we have proved that $d=0$.

As

$\begin{pmatrix}a&1\\\ c&0\end{pmatrix}^{-1}=\begin{pmatrix}0&\frac{1}{c}\\\
1&-\frac{a}{c}\end{pmatrix},$

by considering the inverse map of $M\tau$ and Claim, we get $a=0$. Thus

$M=\begin{pmatrix}0&1\\\ c&0\end{pmatrix}.$

When $n>4$, by applying Lemma 3.8, we see that there exists no $c$. Therefore,
the only possibility of $M$ is of the diagonal form satisfying (33).

Let $n=4$, $s=t=1$. By Lemma 3.9, we derive that the existence of such
$c\in\mathbb{F}_{q^{4}}$ is equivalent to require that
$N_{q^{4}/q^{2}}(\theta)\in\\{N_{q^{4}/q^{2}}(\delta^{\tau}),N_{q^{4}/q^{2}}(1/\delta^{\tau})\\}$
given in Theorem 3.1 (b).

To summary, we have proved that (a) and (b) of Theorem 3.1 are necessary
conditions for the equivalence of $L_{f}$ and $L_{g}$ with $b=0$ and $b\neq
0$. Therefore, the necessity of (a) and (b) in Theorem 3.1 is obtained.

To prove the sufficiency, we only have to let $s=t$. By Lemma 3.6, (a) or (b)
of Theorem 3.1 implies the existence of $d$ such that
$L_{f}=L_{d\overline{g}}$. ∎

It is similar to prove Theorem 3.2 on the $\mathrm{P\Gamma L}$-automorphism
group of a Lunardon-Polverino scattered linear set.

###### Proof of Theorem 3.2.

We only have to follow the proof of Theorem 3.1 under the assumption that
$\theta=\delta$ and $s=t$.

When $b=0$ in (32), we always have $a=1$ and $c=0$. Hence, we just need to
determine $d\in\mathbb{F}_{q^{n}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$ such that $L_{d\overline{f}}=L_{f}$
where $\overline{f}=X^{q^{s}}+\theta^{\tau}X^{q^{n-s}}$. By Lemma 3.6 and
(23), we get the elements in $\mathcal{D}$; see (13). Moreover, by (24),

$|\mathcal{D}|=\begin{cases}N_{\tau}(\theta),&2\nmid n,2\mid q;\\\
2N_{\tau}(\theta),&2\nmid n,2\nmid q;\\\ (q+1)N_{\tau}(\theta),&2\mid
n.\end{cases}$

When $b\neq 0$ in (32), we have $b=1$, $a=0$ and $d=0$. By Lemma 3.8, $n$ must
be $4$. For $n=4$, following the proof of Lemma 3.9, we set $\sigma\colon
x\in\mathbb{F}_{q^{n}}\mapsto x^{q}\in\mathbb{F}_{q^{n}}$,
$h=X^{q}-\theta^{-\tau q^{3}}X^{q^{3}}$, $g=X^{q}+\theta X^{q^{3}}$ and
$d=\frac{c}{\theta^{\tau q}-\theta^{-\tau q^{3}}}$. By Remark 3.7,
$L_{g}=L_{dh}$ if and only if

$d^{q+1}=\left(\frac{\theta}{-\theta^{-\tau q^{3}}}\right)^{q}.$

Hence, we get the elements in $\mathcal{C}$; see (14). Furthermore Lemma 3.9
and Remark 3.7 tell us that

$|\mathcal{C}|=(q+1)N_{\tau}(\theta).$

Therefore, the number of elements in $\mathrm{Aut}(L_{f})$ is the same as in
(15). ∎

##  4\. The number of inequivalent Lunardon-Polverino scattered linear sets

In this section, we determine the total number of inequivalent scattered
polynomials over $\mathbb{F}_{q}$ contained in the Lunardon-Polverino
construction for given $q$. We always assume that $q=p^{r}$ where $p$ is a
prime and $r$ has the following prime factorization

$r=\prod_{i=1}^{\ell}r_{i}^{s_{i}}$

where $r_{1}<r_{2}<\cdots<r_{\ell}$ and $s_{i}\geq 1$ for $i=1,\cdots,\ell$.

Define

$F(r)=\left\\{x\in\mathbb{F}_{p^{r}}:x\notin\mathbb{F}_{p^{r^{\prime}}}\text{
with }r^{\prime}<r\text{ and }r^{\prime}|r\right\\},$

i.e., $F(r)$ consists of the elements in $\mathbb{F}_{q}$ which do not belong
to any proper subfield of $\mathbb{F}_{q}$. Then by the inclusion-exclusion
principle, we get

$|F(r)|=p^{r}-\sum_{i=1}^{\ell}p^{\frac{r}{r_{i}}}+\sum_{\begin{subarray}{c}i,j=1\\\
i<j\end{subarray}}^{\ell}p^{\frac{r}{r_{i}r_{j}}}+\cdots+(-1)^{\ell}p^{\frac{r}{r_{1}r_{2}\cdots
r_{\ell}}}.$

######  Lemma 4.1.

For $0\leq k<r$, if there exists $x\in F(r)$ such that $x^{p^{k}+1}=1$, then
$r=1$ or $r$ is even, and

(51) $k=\begin{cases}0,&r=1;\\\ \frac{r}{2},&2\mid r.\end{cases}$

Moreover, let $K(r)=\left\\{x\in F(r):x^{p^{k}+1}=1,\text{ for some $k$ with
}0\leq k<r\right\\}$. Then

$|K(r)|=\begin{cases}\frac{(-1)^{p+1}+3}{2},&r=1;\\\
p^{\frac{r}{2}}-\frac{(-1)^{p+1}+1}{2},&r=2^{s_{1}};\\\
p^{\frac{r}{2}}-\sum_{i=2}^{\ell}p^{\frac{r}{2r_{i}}}+\sum_{2\leq
i<j\leq\ell}p^{\frac{r}{2r_{i}r_{j}}}+\cdots+(-1)^{\ell-1}p^{\frac{r}{2r_{2}\cdots
r_{\ell}}},&2\mid r\text{ and }\ell>1;\\\ 0,&\text{otherwise}.\end{cases}$

###### Proof.

As $x^{p^{k}+1}=1$, $x^{p^{2k}-1}=1$. Note that $x\in F(r)$, we have
$\gcd(2k,r)=r$. If $2\mid r$, then $k$ must be $\frac{r}{2}$ because $0\leq
k<r$. If $2\nmid r$, then we must have $k=0$ and $r=1$.

Next we determine the size of $K(r)$. If $r=1$, then $k=0$ and
$x^{p^{k}+1}=x^{2}=1$. Thus $|K(r)|=\frac{(-1)^{p+1}+3}{2}$. If $r$ is odd and
$r>1$, then by the first part of the lemma, there is no $x\in F(r)$ such that
$x^{p^{k}+1}=1$. Hence $K(r)$ is empty.

Note that $|\\{x\in\mathbb{F}_{p^{m}}:x^{p^{k}+1}=1\\}|=\gcd(p^{k}+1,p^{m}-1)$
which can be determined by (2). Suppose that
$r=2^{s_{1}}\prod_{i=2}^{\ell}r_{i}^{s_{i}}$ with $s_{1}\geq 1$ which means
$r$ is even. By (51), to determine the size of $K(r)$, we only have to
consider the number of solutions to $x^{p^{k}+1}=1$ for $k=\frac{r}{2}$. If
$\ell=1$ (that is $r$ is a power of $2$), then

$\displaystyle|K(r)|=$ $\displaystyle\left|\left\\{x\in
F(r):x^{p^{r/2}+1}=1\right\\}\right|$ $\displaystyle=$
$\displaystyle\gcd(p^{\frac{r}{2}}+1,p^{r}-1)-\gcd(p^{\frac{r}{2}}+1,p^{\frac{r}{2}}-1)$
$\displaystyle=$ $\displaystyle p^{\frac{r}{2}}-\frac{(-1)^{p+1}+1}{2}.$

For $\ell>1$, by the inclusion-exclusion principle,

$\displaystyle|K(r)|=$ $\displaystyle\left|\left\\{x\in
F(r):x^{p^{r/2}+1}=1\right\\}\right|$ $\displaystyle=$
$\displaystyle\gcd(p^{\frac{r}{2}}+1,p^{r}-1)-\sum_{i=1}^{\ell}\gcd(p^{\frac{r}{2}}+1,p^{\frac{r}{r_{i}}}-1)$
$\displaystyle+\sum_{1\leq
i<j\leq\ell}\gcd(p^{\frac{r}{2}}+1,p^{\frac{r}{r_{i}r_{j}}}-1)+\cdots+(-1)^{\ell}\gcd(p^{\frac{r}{2}}+1,p^{\frac{r}{r_{1}\cdots
r_{\ell}}}-1)$ $\displaystyle=$ $\displaystyle
p^{\frac{r}{2}}+1-\left(\frac{(-1)^{p+1}+3}{2}+\sum_{i=2}^{\ell}\left(p^{\frac{r}{2r_{i}}}+1\right)\right)+\left(\frac{(-1)^{p+1}+3}{2}(\ell-1)+\sum_{2\leq
i<j\leq\ell}\left(p^{\frac{r}{2r_{i}r_{j}}}+1\right)\right)+\cdots$
$\displaystyle+(-1)^{m}\left(\frac{(-1)^{p+1}+3}{2}\binom{\ell-1}{m-1}+\sum_{2\leq
i_{1}<i_{2}<\cdots<i_{m}\leq\ell}\left(p^{\frac{r}{2r_{i_{1}}r_{i_{2}}\cdots
r_{i_{m}}}}+1\right)\right)+\cdots$ $\displaystyle=$ $\displaystyle
p^{\frac{r}{2}}-\sum_{i=2}^{\ell}p^{\frac{r}{2r_{i}}}+\sum_{2\leq
i<j\leq\ell}p^{\frac{r}{2r_{i}r_{j}}}+\cdots+(-1)^{\ell-1}p^{\frac{r}{2r_{2}\cdots
r_{\ell}}}.\qed$

Denote by $\varphi$ the Euler totient function.

######  Theorem 4.2.

For any integer $n\geq 3$ and any prime power $q\neq 2$ with $q=p^{r}$, let
$\Lambda(n,q)$ denote the total number of the $\mathrm{P\Gamma
L}$-inequivalent Lunardon-Polverino scattered linear sets over
$\mathbb{F}_{q^{n}}$. Then

$\Lambda(n,q)=\left(\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}+\epsilon\right)\frac{\varphi(n)}{2}$

if $n$ is odd, and

$\Lambda(n,q)=\left(\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}+\sum_{r^{\prime}\mid
2r,r^{\prime}\nmid
r}\frac{|F(r^{\prime})|-|K(r^{\prime})|}{2r^{\prime}}+\epsilon\right)\frac{\varphi(n)}{2}$

if $n$ is even, where

$\epsilon=\begin{cases}\frac{p-1}{2},&p\neq 2,2\nmid n;\\\
\frac{p-3}{2},&p\neq 2,2\mid n;\\\ 0,&p=2.\end{cases}$

###### Proof.

Recall that a Lunardon-Polverino scattered polynomial is defined as
$f=X^{q^{s}}+\theta X^{q^{n-s}}$.

By Theorem 3.1, for different value of $s$, the scattered linear sets are
inequivalent. Thus we only have to determine the number of polynomials, for
which the associated scattered linear sets are inequivalent, in
$\left\\{X^{q^{s}}+\theta X^{q^{n-s}}:N_{q^{n}/q}(\theta)\neq 0,1\right\\}$
for given $1\leq s<n/2$.

Let

$N(\theta):=\begin{cases}N_{q^{n}/q}(\theta),&2\nmid n;\\\
N_{q^{n}/q^{2}}(\theta),&2\mid n.\end{cases}$

By Theorem 3.1, $X^{q^{s}}+\theta X^{q^{n-s}}$ and $X^{q^{s}}+\delta
X^{q^{n-s}}$ are equivalent if and only if

$N(\theta)=N(\delta)^{p^{k}}\text{ or }1/N(\delta)^{p^{k}}$

for some $k\in\\{0,\cdots,2r-1\\}$ when $n$ is even, and for some
$k\in\\{0,\cdots,r-1\\}$ when $n$ is odd.

Therefore, we have only to count the number of the orbits of
$\mathbb{F}^{*}_{q^{w}}\setminus\\{x\in\mathbb{F}_{q^{w}}:N_{q^{w}/q}(x)=1\\}$
under the action of the group $G$ generated by
$\mathrm{Aut}(\mathbb{F}_{q^{w}})$ and the map
$x\in\mathbb{F}_{q^{w}}^{*}\mapsto\frac{1}{x}\in\mathbb{F}_{q^{w}}^{*}$, where
$w=1$ for odd $n$ and $w=2$ for even $n$. Note that

$G=\\{x\in\mathbb{F}_{q^{w}}^{*}\mapsto x^{\pm
p^{k}}\in\mathbb{F}_{q^{w}}^{*}:k\in\\{0,\cdots,wr-1\\}\\}.$

Depending on the parity of $n$, we separate the proof into two cases.

Case (I). $2\nmid n$. Note that
$\mathbb{F}_{q}\setminus\\{0,1\\}=\left(\bigcup_{r^{\prime}\mid
r,r^{\prime}\neq
1}F(r^{\prime})\right)\cup\left(F(1)\setminus\\{0,1\\}\right)$. When
$r^{\prime}$ is odd, by Lemma 4.1, $|K(r^{\prime})|=0$ which means there is no
$x\in F(r^{\prime})$ such that $x^{p^{k}}=1/x$ for any $k$. Consequently,
there are exactly $\frac{|F(r^{\prime})|}{2r^{\prime}}$ orbits of
$F(r^{\prime})$ under $G$ for $r^{\prime}>1$. For $r^{\prime}=1$, it is easy
to see that the elements in $F(1)\setminus\\{0,1\\}$ is partitioned into
$\epsilon=\frac{p-1}{2}$ orbits for $p$ odd, and for $p=2$ there is no element
in $F(1)\setminus\\{0,1\\}$.

When $r^{\prime}$ is even, suppose that there are $N_{1}$ orbits of length
$2r^{\prime}$ and $N_{2}$ orbits of length $r^{\prime}$ of $F(r^{\prime})$
under $G$. Then

$2r^{\prime}N_{1}+r^{\prime}N_{2}=|F(r^{\prime})|.$

By the definition of $K(r^{\prime})$,
$N_{2}=\frac{|K(r^{\prime})|}{r^{\prime}}$. By simple computation,

$N_{1}=\frac{|F(r^{\prime})|-|K(r^{\prime})|}{2r^{\prime}}~{}~{}\text{ and
}~{}~{}N_{1}+N_{2}=\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}.$

Since $|K(r^{\prime})|=0$ for odd $r^{\prime}>1$, the total number of orbits
of elements in $\mathbb{F}_{q}\setminus\\{0,1\\}$ are

$\sum_{r^{\prime}\mid r,r^{\prime}\neq
1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}+\epsilon.$

Case (II). $2\mid n$. Note that

$\mathbb{F}^{*}_{q^{2}}\setminus\\{x\in\mathbb{F}_{q^{2}}:x^{q+1}=1\\}=\bigcup_{r^{\prime}\mid
2r}\hat{F}(r^{\prime})=\left(\bigcup_{r^{\prime}\mid 2r,r^{\prime}\neq
1}\hat{F}(r^{\prime})\right)\cup\left(F(1)\setminus\\{0,\pm 1\\}\right),$

where $\hat{F}(r^{\prime}):=F(r^{\prime})\setminus\\{x\in
F(r^{\prime}):x^{q+1}\in\\{0,1\\}\\}$.

When $r^{\prime}\mid r$, $x^{q+1}=x^{2}$. Hence, if $r^{\prime}>1$, then
$\hat{F}(r^{\prime})=F(r^{\prime})$.

When $r^{\prime}\mid 2r$ and $r^{\prime}\nmid r$, $r^{\prime}$ must be even
and $x^{q+1}=x^{p^{r^{\prime}/2}+1}$. By Lemma 4.1,
$\hat{F}(r^{\prime})=F(r^{\prime})\setminus K(r^{\prime})$. To summarize, for
$r^{\prime}\mid 2r$,

(52) $\hat{F}(r^{\prime})=\begin{cases}F(r^{\prime}),&r^{\prime}\mid r;\\\
F(r^{\prime})\setminus K(r^{\prime}),&r^{\prime}\nmid r.\end{cases}$

By the above analysis and the counting argument in Case (I), when $r^{\prime}$
is odd and $r^{\prime}>1$, there are exactly
$\frac{|F(r^{\prime})|}{2r^{\prime}}$ orbits of the elements in
$\hat{F}(r^{\prime})$ under $G$.

When $r^{\prime}$ is even, we have to take care of two subcases. If
$r^{\prime}\mid r$, then by the counting argument in Case (I), there are
exactly $\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}$ orbits of the
elements in $\hat{F}(r^{\prime})$ under $G$. If $r^{\prime}\nmid r$, the
number of orbits is $N_{1}$ which equals
$\frac{|F(r^{\prime})|-|K(r^{\prime})|}{2r^{\prime}}$.

Therefore, the total number of orbits is

$\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}+\sum_{r^{\prime}\mid
2r,r^{\prime}\nmid
r}\frac{|F(r^{\prime})|-|K(r^{\prime})|}{2r^{\prime}}+\epsilon,$

where $\epsilon$ equals the number of the orbits in $\hat{F}(1)$ under the
action of $G$. ∎

When $r=1$, the value of $\Lambda(n,q)$ is $\epsilon\frac{\varphi(n)}{2}$ by
Theorem 4.2 (where $\epsilon$ is as in Theorem 4.2). However, for $r$ with
many divisors, it is in general not trivial to see the explicit value of
$\Lambda(n,q)$. To conclude this section, we provide an upper bound and a
lower bound for it and we discuss about its asymptotics.

######  Theorem 4.3.

For any integer $n\geq 3$ and any prime power $q\neq 2$ with $q=p^{r}$ and
$r>1$, let $\Lambda(n,q)$ denote the total number of the $\mathrm{P\Gamma
L}$-inequivalent Lunardon-Polverino scattered linear sets over
$\mathbb{F}_{q^{n}}$. Let $\sigma(r)$ denote the sum of divisors of $r$, i.e.
$\sigma(r)=\sum_{d\mid r}d$. When $n$ is odd,

(53)
$\frac{p^{r}-p}{2r}<\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon<\begin{cases}\frac{p^{r}}{2r}\left(1+\frac{\sigma(r)-r-1}{p^{r(1-\frac{1}{r_{1}})}}\right),&2\nmid
r;\\\
\frac{p^{r}}{2r}\left(1+\frac{\sigma(r)-r-1}{p^{\frac{r}{2}}}\right)+\sum_{i=1}^{s_{1}}p^{2^{i-1}},&r=2^{s_{1}};\\\
\frac{p^{r}}{2r}\left(1+\frac{\sigma(r)-r}{p^{\frac{r}{2}}}+\frac{\sigma(r)-r-1}{p^{\frac{3r}{4}}}\right),&2\mid
r\text{ and }\ell>1.\end{cases}$

When $n$ is even,

(54)
$\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon<\begin{cases}\frac{p^{2r}}{4r}\left(1+\frac{\sigma(2r)-2r-1}{p^{r}}\right),&2\nmid
r;\\\
\frac{p^{2r}}{4r}\left(1+\frac{\sigma(2r)-2r-1}{p^{r}}\right)+\sum_{i=1}^{s_{1}}p^{2^{i-1}},&r=2^{s_{1}};\\\
\frac{p^{2r}}{4r}\left(1+\frac{\sigma(2r)-2r-1}{p^{r}}+\frac{2}{p^{\frac{3r}{2}}}+\frac{2(\sigma(r)-r-1)}{p^{\frac{7r}{4}}}\right),&2\mid
r\text{ and }\ell>1.\end{cases}$

and

(55)
$\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon>\begin{cases}\frac{p^{2r}-p}{4r}-\sum_{i=1}^{s_{1}+1}p^{2^{i-1}},&r=2^{s_{1}};\\\
\frac{p^{2r}}{4r}\left(1-\frac{1}{p^{r}}-\frac{\sigma(2r)-2r-1}{p^{\frac{3r}{2}}}-\frac{1}{p^{2r-1}}\right),&\text{otherwise}.\end{cases}$

According to a classical result by Gronwall [10], the value of

$\limsup_{r\rightarrow\infty}\frac{\sigma(r)}{r\log\log r}=e^{\gamma}$

where $\gamma=0.5772156\dots$ denotes the Euler-Mascheroni constant.
Therefore, by Theorem 4.2, for given $n$, the value of $\Lambda(n,q)$ is
approximately $\frac{p^{r}\varphi(n)}{4r}$ for odd $n$ and
$\frac{p^{2r}\varphi(n)}{8r}$ for even $n$ provided that $p^{r}$ is large
enough.

###### Proof of Theorem 4.3.

We start by determining the upper bounds. To this aim we first give upper
bounds on the quantities appearing in the expression of $\Lambda(n,q)$ as in
Theorem 4.2.

As in the very beginning of Section 4, we set $r_{1}$ to be the smallest prime
divisor of $r$. As $F(m)\subsetneq\mathbb{F}_{p^{m}}$ for any $m>1$,

$\displaystyle\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|}{2r^{\prime}}=$
$\displaystyle\frac{1}{2r}\sum_{r^{\prime}\mid
r,r^{\prime}<r}r^{\prime}\left|F\left(\frac{r}{r^{\prime}}\right)\right|$
$\displaystyle<$ $\displaystyle\frac{1}{2r}\left(p^{r}+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}r^{\prime}p^{\frac{r}{r^{\prime}}}\right)$ $\displaystyle=$
$\displaystyle\frac{p^{r}}{2r}\left(1+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}\frac{r^{\prime}}{p^{r-\frac{r}{r^{\prime}}}}\right)$
$\displaystyle<$ $\displaystyle\frac{p^{r}}{2r}\left(1+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}\frac{r^{\prime}}{p^{r(1-1/r_{1})}}\right).$

Thus

(56) $\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|}{2r^{\prime}}<\frac{p^{r}}{2r}\left(1+\frac{\sigma(r)-r-1}{p^{r(1-1/r_{1})}}\right).$

Next we derive an upper bound on the value of $\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}$. Following the notation
and results given in Lemma 4.1, if $2\nmid r$ and $r>1$, then

(57) $\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}=0.$

If $r=2^{s_{1}}$ and $r>1$, then

(58) $\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}=\sum_{i=1}^{s_{1}}\left(p^{2^{i-1}}-\frac{(-1)^{p+1}+1}{2}\right)\leq\sum_{i=1}^{s_{1}}p^{2^{i-1}},$

since $|K(r^{\prime})|\leq p^{\frac{r^{\prime}}{2}}$. If $2\mid r$ and
$\ell>1$, then

$\displaystyle\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}=$
$\displaystyle\frac{1}{2r}\sum_{r^{\prime}\mid
r,r^{\prime}<r}r^{\prime}\left|K\left(\frac{r}{r^{\prime}}\right)\right|$
$\displaystyle<$
$\displaystyle\frac{p^{\frac{r}{2}}}{2r}\left(1+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}\frac{r^{\prime}}{p^{\frac{r}{2}-\frac{r}{2r^{\prime}}}}\right)$
$\displaystyle<$
$\displaystyle\frac{p^{\frac{r}{2}}}{2r}\left(1+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}\frac{r^{\prime}}{p^{\frac{r}{4}}}\right),$

where the first inequality is obtained using that

$\sum_{r^{\prime}\mid
r,r^{\prime}<r}r^{\prime}\left|K\left(\frac{r}{r^{\prime}}\right)\right|=\left|K\left(r\right)\right|+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}r^{\prime}\left|K\left(\frac{r}{r^{\prime}}\right)\right|<p^{\frac{r}{2}}+\sum_{r^{\prime}\mid
r,1<r^{\prime}<r}r^{\prime}p^{\frac{r}{2r^{\prime}}}.$

Hence,

(59) $\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}<\frac{p^{\frac{r}{2}}}{2r}\left(1+\frac{\sigma(r)-r-1}{p^{\frac{r}{4}}}\right).$

Now, we are ready to prove the upper bound of $\Lambda(n,q)$.

When $n$ is odd, plugging (56), (57), (58) and (59) into

$\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon=\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}},$

we get the upper bound given in (53).

When $n$ is even,

$\displaystyle\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon=$
$\displaystyle\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r^{\prime}}+\sum_{r^{\prime}\mid
2r,r^{\prime}\nmid r}\frac{|F(r^{\prime})|-|K(r^{\prime})|}{2r^{\prime}}$
$\displaystyle<$ $\displaystyle\sum_{r^{\prime}\mid
2r,r^{\prime}>1}\frac{|F(r^{\prime})|}{2r^{\prime}}+\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}.$

Plugging (56), (57), (58) and (59) into it, we get (54).

Finally, we compute the lower bound. First, it is easy to see that

$\sum_{r^{\prime}\mid
r,r^{\prime}>1}|F(r^{\prime})|=|\mathbb{F}_{p^{r}}\setminus\mathbb{F}_{p}|=p^{r}-p.$

When $n$ is odd,

$\displaystyle\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon>$
$\displaystyle\sum_{r^{\prime}\mid
r,r^{\prime}>1}\frac{|F(r^{\prime})|+|K(r^{\prime})|}{2r}$ $\displaystyle>$
$\displaystyle\frac{p^{r}-p}{2r}.$

Therefore, (53) is proved.

When $n$ is even,

$\displaystyle\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon>$
$\displaystyle\sum_{r^{\prime}\mid
2r,r^{\prime}>1}\frac{|F(r^{\prime})|}{2r^{\prime}}-\sum_{r^{\prime}\mid
2r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}$ $\displaystyle>$
$\displaystyle\frac{p^{2r}-p}{4r}-\sum_{r^{\prime}\mid
2r,r^{\prime}>1}\frac{|K(r^{\prime})|}{2r^{\prime}}.$

Then

$\frac{\Lambda(n,q)}{\varphi(n)/2}-\epsilon>\begin{cases}\frac{p^{2r}-p}{4r}-\sum_{i=1}^{s_{1}+1}p^{2^{i-1}},&r=2^{s_{1}};\\\
\frac{p^{2r}-p}{4r}-\frac{p^{r}}{4r}\left(1+\frac{\sigma(2r)-2r-1}{p^{\frac{r}{2}}}\right),&\text{otherwise}.\end{cases}$

Consequently, we obtain (55). ∎

## Acknowledgment

The authors express their gratitude to the anonymous reviewers for
constructive comments which are helpful to the improvement of the presentation
of this paper. Wei Tang and Yue Zhou were supported by the Sino-German
Mobility Programme M-0157 and the Training Program for Excellent Young
Innovators of Changsha (No. kq2106006). The research of Ferdinando Zullo was
supported by the project “VALERE: VAnviteLli pEr la RicErca” of the University
of Campania “Luigi Vanvitelli” and was partially supported by the Italian
National Group for Algebraic and Geometric Structures and their Applications
(GNSAGA - INdAM).

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# Gauge Boson Mass Dependence and Chiral Anomalies in Generalized Massless
Schwinger models

Howard Georgi
Center for the Fundamental Laws of Nature
Jefferson Physical Laboratory
Harvard University
Cambridge, MA 02138<EMAIL_ADDRESS>
###### Abstract

I bosonize the position-space correlators of flavor-diagonal scalar fermion
bilinears in arbitrary generalizations of the Schwinger model with $n_{F}$
massless fermions coupled to $n_{A}$ gauge bosons for $n_{F}\geq n_{A}$. For
$n_{A}=n_{F}$, the fermion bilinears can be bosonized in terms of $n_{F}$
scalars with masses proportional to the gauge couplings. As in the Schwinger
model, bosonization can be used to find all correlators, including those that
are forbidden in perturbation theory by anomalous chiral symmetries, but there
are subtleties when there is more than one gauge boson. The new result here is
the general treatment of the dependence on gauge boson masses in models with
more than one gauge symmetry. For $n_{A}<n_{F}$, there are fermion bilinears
with nontrivial anomalous dimensions and there are unbroken chiral symmetries
so some correlators vanish while others are non-zero due to chiral anomlies.
Taking careful account of the dependence on the masses, I show how the
$n_{A}<n_{F}$ models emerge from $n_{A}=n_{F}$ as gauge couplings (and thus
gauge boson masses) go to zero. When this is done properly, the limit of zero
gauge coupling is smooth. Our consistent treatment of gauge boson masses
guarantees that anomalous symmetries are broken while unbroken chiral
symmetries are preserved because correlators that break the non-anomalous
symmetries go to zero in the limit of zero gauge coupling..

##  1 Introduction

In this note, I analyze the position-space correlators of flavor-diagonal
scalar fermion bilinears in arbitrary generalizations of the massless
Schwinger model and focus on the behavior of gauge anomalies in the limit in
which some gauge couplings go to zero. We will see that when the dimensional
gauge couplings are properly accounted for (which I believe is done here for
the first time) the limit is smooth.

We consider a general Schwinger model in 1+1 dimensions with $n_{F}$ massless
Dirac fermions, $\psi_{\alpha}$ for $\alpha=1$ to $n_{F}$, and $n_{A}\leq
n_{F}$ vector bosons, $A^{\mu}_{j}$ for $j=1$ to $n_{A}$.

The Lagrangian is (using summation convention where it does not cause
confusion)

$\mathcal{L}_{Af}=\left(\sum_{\alpha=1}^{n_{F}}\overline{\psi}_{\alpha}\,\left(i\\!\not\\!\partial\,-e_{\alpha
j}\not\\!\\!A_{j}\right)\,\psi_{\alpha}\right)-\frac{1}{4}F_{j}^{\mu\nu}F_{j\mu\nu}$
(1)

Note that we have assumed that the gauge couplings are diagonal in the fermion
space. This is important to ensure that the model is exactly solvable. We
could have non-diagonal couplings as long as the couplings to different gauge
bosons commute with one another. But then we can simultaneously diagonalize
them with a unitary transformation on the fermion fields, so we will assume
that the gauge bosons only couple to diagonal fermion currents. The gauge
couplings are an $n_{F}\times n_{A}$ matrix, $e$ in which

$e_{\alpha j}\mbox{~{}is the coupling of the $j$th vector to the $\alpha$th
diagonal fermion current.}$ (2)

If $n_{F}<n_{A}$, there are linear combinations of the gauge bosons that do
not couple to anything and can be safely ignored, so we will always assume
that $n_{F}\geq n_{A}$.

Each of the massless fermions generates a contribution to the vector boson
mass matrix [1] and we will use our freedom to redefine the vector fields to
diagonalize the physical vector boson mass so we can write

$\sum_{\alpha}e_{\alpha j}e_{\alpha
j^{\prime}}=\pi\,m_{j}^{2}\,\delta_{jj^{\prime}}\quad\quad\quad
m_{j}^{2}=\frac{1}{\pi}\sum_{\alpha}e^{2}_{\alpha j}$ (3)

Then the $m_{j}$ are the physical gauge boson masses. The basis of the
fermions is fixed up to unbroken flavor symmetries because the currents must
be diagonal. And unless some gauge boson masses are equal, the basis of the
gauge boson fields is also fixed by (3).

We work in Lorenz gauge,

$\partial_{\mu}A_{j}^{\mu}=0$ (4)

Then a generalization of the arguments of [1]111For details and definitions,
see [2]. The conjecture in the paper was shown to be false in [3] but we can
still use the calculational tools. shows that the gauge invariant correlation
functions to all orders in perturbation theory can be found using the free-
field Lagrangian describing massless fermions, $\Psi_{\alpha}$, massive
bosons, $\mathcal{B}_{j}$, and massless scalar ghosts, $\mathcal{C}_{j}$

$\mathcal{L}=\left(\sum_{\alpha=1}^{n_{F}}i\overline{\Psi}_{\alpha}\\!\not\\!\partial\,\Psi_{\alpha}\right)-\frac{m_{j}^{2}}{2}\mathcal{B}_{j}^{2}+\frac{1}{2}\partial_{\mu}\mathcal{B}_{j}\partial^{\mu}\mathcal{B}_{j}-\frac{1}{2}\partial_{\mu}\mathcal{C}_{j}\partial^{\mu}\mathcal{C}_{j}$
(5)

using the replacements

$A_{j}^{\mu}=\epsilon^{\mu\nu}\partial_{\nu}\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)/m_{j}$
(6) $\psi_{\alpha}=e^{-i\left(\pi\right)^{1/2}\,\left(e_{\alpha
j}/m_{j}\right)\left(\mathcal{C}_{j}-\mathcal{B}_{j}\right)\gamma^{5}}\Psi_{\alpha}$
(7)

The massive and the massless free scalar propagators, respectively,
are222$K_{0}$ is the modified Bessel function of the second kind.

$-i\braket{0}{T\mathcal{B}_{j}\left(x\right)\mathcal{B}_{k}\left(0\right)}{0}=\delta_{jk}\int\frac{d^{2}p}{\left(2\pi\right)^{2}}\frac{e^{-ipx}}{p^{2}-m_{j}^{2}+i\epsilon}=-\delta_{jk}\frac{i}{2\pi}K_{0}\left(m_{j}\sqrt{-x^{2}+i\epsilon}\right)$
(8)
$-i\braket{0}{T\mathcal{C}_{j}\left(x\right)\mathcal{C}_{k}\left(0\right)}{0}=-\delta_{jk}\int\frac{d^{2}p}{\left(2\pi\right)^{2}}\frac{e^{-ipx}}{p^{2}+i\epsilon}=-\delta_{jk}\frac{i}{2\pi}\ln\left(\xi
m_{j}\sqrt{-x^{2}+i\epsilon}\right)$ (9) $\xi\equiv
e^{\gamma_{E}}/2\mbox{~{}~{}where $\gamma_{E}$ is Euler's constant}$ (10)

The arbitrary dimensional constant in the logarithmic ghost propagators has
been fixed so that the $\mathcal{B}_{j}$ and $\mathcal{C}_{j}$ propagators
exactly cancel when the gauge couplings vanish, $m_{j}=0$.

We will focus on the position-space correlators of the flavor-diagonal
fermion-bilinear scalar operators

${\begin{array}[]{l}\displaystyle
O_{\alpha+}\equiv\psi_{2\alpha}^{*}\,\psi_{1\alpha}=e^{-2i\sum_{j}\left(e_{\alpha
j}/m_{j}\right)\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)}\Psi^{*}_{2\alpha}\Psi_{1\alpha}\\\
\displaystyle
O_{\alpha-}\equiv\psi_{1\alpha}^{*}\,\psi_{2\alpha}=e^{2i\sum_{j}\left(e_{\alpha
j}/m_{j}\right)\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)}\Psi^{*}_{1\alpha}\Psi_{2\alpha}=O_{\alpha+}^{*}\end{array}}$
(11)

While (11) gives a complete description of the $O_{\alpha\pm}$ correlators to
all orders in perturbation theory, the nonperturbative effects of gauge
anomalies are much more transparent if we bosonize. For massless free fermions
in 1+1, any non-zero correlator of diagonal fermion bilinears can be
calculated by replacing the bilinears with exponentials of free massless
scalar fields according to

$\Psi^{*}_{2\alpha}\Psi_{1\alpha}\to\frac{\xi\mathcal{M}_{\alpha}}{2\pi}\,e^{-2i\pi^{1/2}\mathcal{D}^{\alpha}}\quad\quad\Psi^{*}_{1\alpha}\Psi_{2\alpha}\to\frac{\xi\mathcal{M}_{\alpha}}{2\pi}\,e^{2i\pi^{1/2}\mathcal{D}^{\alpha}}$
(12)

The $\mathcal{M}_{\alpha}$s are arbitrary and correlated with the propagators
of the bosonization fields, $\mathcal{D}^{\alpha}$.

$-i\braket{0}{T\mathcal{D}^{\alpha}\left(x\right)\mathcal{D}^{\beta}\left(0\right)}{0}=\delta^{\alpha\beta}\int\frac{d^{2}p}{\left(2\pi\right)^{2}}\frac{e^{-ipx}}{p^{2}+i\epsilon}=\delta^{\alpha\beta}\frac{i}{2\pi}\ln\left(\xi\mathcal{M}_{\alpha}\sqrt{-x^{2}+i\epsilon}\right)$
(13)

There is no mass in the position space correlators of the fermion bilinears
and the mass $\mathcal{M}_{\alpha}$ is introduced by the bosonization
procedure which requires a mass to get the dimensions right. (12) is a
straightforward consequence of Fermi-Dirac statistics. Then

$O_{\alpha\pm}=\frac{\xi\mathcal{M}_{\alpha}}{2\pi}\,e^{\mp
2i\sum_{j=1}^{n_{A}}\left(e_{\alpha
j}/m_{j}\right)\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)}\,e^{\mp
2i\pi^{1/2}\mathcal{D}^{\alpha}}$ (14)

The key to using (14) most effectively will be to cancel the ghost fields with
linear combinations of the bosonization fields. This will be easier if we
change notation slightly.

First, we define an orthogonal $n_{F}\times n_{F}$ matrix $\eta_{\alpha j}$
where the $\alpha$ index labels the fermion as usual but the $j$ index is
extended to make the matrix square and a subset $b$ of the $j$ indices are
associated with the gauge fields.

$\eta\,\eta^{T}=I\quad\quad\eta_{\alpha j}=\frac{e_{\alpha
j}}{m_{j}\sqrt{\pi}}\mbox{~{}~{}for~{}~{}$j\in b$}$ (15)

we can write

$O_{\alpha\pm}=\frac{\xi\mathcal{M}_{\alpha}}{2\pi}\,e^{\mp
2i\pi^{1/2}\sum_{j\in b}\eta_{\alpha
j}\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)}\,e^{\mp
2i\pi^{1/2}\mathcal{D}^{\alpha}}$ (16)

Now we can define linear combinations of the bosonization fields

$\mathcal{D}_{j}\equiv\eta_{\alpha j}\mathcal{D}^{\alpha}$ (17)

and write

$O_{\alpha\pm}=\frac{\xi\mathcal{M}_{\alpha}}{2\pi}\,e^{\mp
2i\pi^{1/2}\sum_{j\in b}\eta_{\alpha
j}\left(\mathcal{B}_{j}-\mathcal{C}_{j}\right)}\,e^{\mp
2i\pi^{1/2}\sum_{j}\eta_{\alpha j}\mathcal{D}_{j}}$ (18)

In this form, it looks like we can cancel the $\mathcal{D}_{j}$ fields for
$j\in b$ with the ghosts. That is precisely what happens in the original
Schwinger model and in the Schwinger model with flavors. But in general, this
cancellation is not exact because of the different masses associated with
different gauge couplings.

The $\mathcal{D}_{j}$ propagator is

$-i\braket{0}{T\mathcal{D}_{j}\left(x\right)\mathcal{D}_{k}\left(0\right)}{0}=\sum_{\alpha}\eta_{j\alpha}\eta_{k\alpha}\frac{i}{2\pi}\ln\left(\xi\mathcal{M}_{\alpha}\sqrt{-x^{2}+i\epsilon}\right)$
(19)

Thus for the $\mathcal{D}_{j}$ for $j=1$ to $n_{A}$ to exactly cancel the
ghosts $\mathcal{C}_{j}$ we must have

$\sum_{\alpha}\eta_{\alpha j}\eta_{\alpha
k}\log\mathcal{M}_{\alpha}=\delta_{jk}\log m_{j}\mbox{~{}~{}~{}for $j$ or
$k\in b$}$ (20)

and (18) will be most useful if the $\mathcal{D}_{j}$ propagator is diagonal
for all $j$. In general, this is impossible and the most useful thing we can
do is to choose a common bosonization mass $\mathcal{M}_{\alpha}=m$ for the
$\mathcal{D}^{\alpha}$s:

$\mathcal{M}_{\alpha}=m\mbox{~{}~{}~{}for all $\alpha$.}$ (21)

so that

$-i\braket{0}{T\mathcal{D}_{j}\left(x\right)\mathcal{D}_{k}\left(0\right)}{0}=\delta_{jk}\frac{i}{2\pi}\ln\left(\xi
m\sqrt{-x^{2}+i\epsilon}\right)$ (22)

We could always cancel the ghost for a single $j$ by choosing $m=m_{j}$. This
is the way the issue is typically handled in models with a single gauge boson
but it is not adequate for $n_{A}>1$.

If we adopt (22), we can almost cancel the ghosts in general.

$O_{\alpha\pm}=\frac{\xi m}{2\pi}\,e^{\mp 2i\pi^{1/2}\sum_{j\in b}\eta_{\alpha
j}\left(\mathcal{B}_{j}-\mathcal{E}_{j}\right)}\,e^{\mp 2i\pi^{1/2}\sum_{j\in
b^{\prime}}\eta_{\alpha j}\mathcal{D}_{j}}$ (23)

where $b^{\prime}$ is the complement of $b$ and the $\mathcal{E}_{j}$ are
“constant” fields

$\mathcal{E}_{j}=\mathcal{C}_{j}-\mathcal{D}_{j}\mbox{~{}~{}~{}for $j\in b$}$
(24)

with propagators

$-i\braket{0}{T\mathcal{E}_{j}\left(x\right)\mathcal{E}_{k}\left(0\right)}{0}=\delta_{jk}\frac{i}{2\pi}\log\left(m-m_{j}\right)$
(25)

so the Wick expansion gives

$\braket{0}{Te^{2i\pi^{1/2}\mathcal{E}_{j}\left(x\right)}\,e^{\mp
2i\pi^{1/2}\mathcal{E}_{j}\left(0\right)}}{0}=\left(m_{j}/m\right)^{\pm 2}$
(26)

The $\mathcal{E}_{j}$ fields keep the engineering dimensions right while
eliminating the dependence on the arbitrary bosonization mass $m$. (23-25) can
be used to calculate any matrix element that is non-zero in perturbation
theory and we can simplify (23) further by eliminating the $\mathcal{E}_{j}$
operators.333Note that we will soon use (23) to calculate other matrix
elements, but because these more general results follow from the perturbative
calculation and cluster decomposition, the simplification will work in
general.

Look at a general perturbatively non-zero correlator involving only bilinears
(no higher powers). If there are $n_{\alpha}$ $O_{\alpha+}$s there must also
be $n_{\alpha}$ $O_{\alpha-}$s or the correlator would vanish in perturbation
theory. We can look separately at the contributions from each
$\mathcal{E}_{j}$ because the propagators don’t mix. So we are interested in

$\braket{0}{T\prod_{r=1}^{n_{\alpha}}e^{2i\pi^{1/2}\eta_{\alpha
j}\mathcal{E}_{j}\left(x_{r}\right)}\,e^{-2i\pi^{1/2}\eta_{\alpha
j}\mathcal{E}_{j}\left(y_{r}\right)}}{0}$ (27)

We have put arguments $\left(x_{r}\right)$ and $\left(y_{r}\right)$ on the
$\mathcal{E}_{j}$ from $O_{\alpha+}\left(x_{r}\right)$ and
$O_{\alpha-}\left(y_{r}\right)$ for notational convenience, but the
$\mathcal{E}_{j}$ are constant operators so the Wick expansion of (27) is
independent of the coordinates. The contribution from contractions of the
$\mathcal{E}_{j}\left(x_{r}\right)$s with the
$\mathcal{E}_{j}\left(y_{r}\right)$s is

$\left(\frac{m_{j}^{2}}{m^{2}}\right)^{n_{\alpha}^{2}\eta_{\alpha j}^{2}}$
(28)

The contribution from contractions of the $\mathcal{E}_{j}\left(x_{r}\right)$s
with $\mathcal{E}_{j}\left(x^{\prime}_{r}\right)$s is

$\left(\frac{m_{j}^{2}}{m^{2}}\right)^{-\left(n_{\alpha}^{2}-n_{\alpha}\right)\eta_{\alpha
j}^{2}/2}$ (29)

as is the contribution from contractions of the
$\mathcal{E}_{j}\left(y_{r}\right)$s with
$\mathcal{E}_{j}\left(y^{\prime}_{r}\right)$s. So the total contribution is

$\left(\frac{m_{j}^{2}}{m^{2}}\right)^{n_{\alpha}\eta_{\alpha j}^{2}}$ (30)

This means that we can associate a factor

$N_{\alpha}=\prod_{j\in b}\left(\frac{m_{j}}{m}\right)^{\eta_{\alpha j}^{2}}$
(31)

with each operator in each $O_{\alpha+}$-$O_{\alpha-}$ pair and completely
capture the effects of the $\mathcal{E}_{j}$s for perturbatively non-zero
correlators.

Now we can eliminate the $\mathcal{E}_{j}$s and write

${\begin{array}[]{c}\displaystyle O_{\alpha\pm}=\frac{\xi m}{2\pi}\,e^{2\pm
i\pi^{1/2}\sum_{j\in b}\eta_{\alpha
j}\left(\mathcal{E}_{j}-\mathcal{B}_{j}\right)}\,e^{2\mp i\pi^{1/2}\sum_{j\in
b^{\prime}}\eta_{\alpha j}\mathcal{D}_{j}}\to\\\ \displaystyle\frac{\xi
m}{2\pi}\,N_{\alpha}\,e^{-2\pm i\pi^{1/2}\left(\sum_{j\in b}\eta_{\alpha
j}\mathcal{B}_{j}+\sum_{j\in b^{\prime}}\eta_{\alpha
j}\mathcal{D}_{j}\right)}\end{array}}$ (32)

We can use (32) to calculate any correlator that is non-zero in perturbation
theory. A particularly nice feature of (32) is the way it encodes the diagonal
chiral symmetries. The theory is invariant under global translations of the
massless $\mathcal{D}_{j}$ fields for $j\in b^{\prime}$. These translations
generate the $n_{F}-n_{A}$ unbroken chiral symmetries on the fermion
bilinears.

$O_{\alpha\pm}$ has $j$ chiral charge $q_{j}=\pm\eta_{\alpha j}$ for
$\delta\mathcal{D}_{j}=\theta_{j}$ with $j\in b^{\prime}$. (33)

There are no unbroken chiral transformations from translations of
$\mathcal{D}_{j}$ for $j\in b$, because these $\mathcal{D}_{j}$s have been
eaten by the ghosts. The corresponding chiral transformation from translations
of the $\mathcal{B}_{j}$ for $j\in b$ are softly broken by the
$\mathcal{B}_{j}$ mass terms generated by the gauge anomalies.

##  2 Cluster Decomposition

While we only calculate to all orders in perturbation theory, in some
situations, cluster decomposition gives us nonperturbative information. For
any combination of $n_{o}$ fermion bilinears in some region of space-time, we
can look at a correlator that also includes all the conjugate fields in a
region far away in space as in

$\Braket{0}{T\,\prod_{u=1}^{n_{o}}O_{\alpha_{u}s_{u}}(x_{u})\,O_{\alpha_{u}s_{u}}^{*}(z+y_{u})}{0}$
(34)

where the $s_{u}$ are $\pm 1$, as in (11) and the $x_{a}$ and $y_{a}$ are
clustered in some region of size $\ell$ ($-(x_{a}-x_{b})^{2}<\ell^{2}$) around
the (arbitrary) origin and $z$ is a large space-like 2-vector. This correlator
is calculable in perturbation theory. Using (32), we can write it as

$\left(\frac{\xi
m}{2\pi}\right)^{2n_{o}}\left(\prod_{u}N_{\alpha_{u}}\right)\,\left(X\,Y\,Z\right)$
(35)

where

${\begin{array}[]{c}X=\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\Delta_{j}(x_{u}-x_{u^{\prime}})\right]\\\
Y=\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\Delta_{j}(y_{u}-y_{u^{\prime}})\right]\\\
Z=\prod_{u,u^{\prime}}\exp\left[2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\Delta_{j}(z+y_{u^{\prime}}-x_{u})\right]\end{array}}$
(36)

where

$\Delta_{j}(x)=\left\\{{\begin{array}[]{ll}K_{0}\left(m_{j}\sqrt{-x^{2}+i\epsilon}\right)&\mbox{for
$j\in b$}\\\ -\log\left(m\sqrt{-x^{2}+i\epsilon}\right)&\mbox{for $j\in
b^{\prime}$}\end{array}}\right.$ (37)

We can now study the correlator as $-z^{2}$ goes to infinity.

If $b^{\prime}\neq\emptyset$, that is for $n_{F}>n_{A}$ the correlator will in
general go to zero, with the $Z$ factor falling off like

$\frac{1}{\left(m\sqrt{-x^{2}+i\epsilon}\right)^{2\sum_{j\in
b^{\prime}}Q_{j}^{2}}}$ (38)

where $Q_{j}$ is the total $j$ chiral charge of the operators in the $X$
region,

$Q_{j}=\sum_{u}s_{u}\eta_{\alpha_{u}j}$ (39)

Thus $\sum_{j\in b^{\prime}}Q_{j}^{2}$ is the anomalous dimension of the
leading operators in the operator product expansion of the set
$\\{O_{\alpha_{u}s_{u}}(x_{u})\\}$ and I will refer this as the anomalous
dimension of the set and we will have more to say about it below.

But we will begin by assuming that $n_{A}=n_{F}$ so there are no massless
fields ($b^{\prime}=\emptyset$) and

$\Delta_{j}(x)=K_{0}\left(m_{j}\sqrt{-x^{2}+i\epsilon}\right)\;\forall\;j$
(40)

Then for sufficiently large $z$, all the Bessel functions go to $0$ and $Z\to
1$. If the correlator is non-zero as the regions move infinitely far apart, it
must factor into a product of contributions in the two regions.

$\Braket{0}{T\,\prod_{u=1}^{n_{o}}O_{\alpha_{u}s_{u}}(x_{u})\,O_{\alpha_{u}s_{u}}^{*}(z+y_{u})}{0}\mathop{\longrightarrow}\limits_{-z^{2}\to\infty}\left(\frac{\xi
m}{2\pi}\right)^{2n_{o}}\left(\prod_{u}N_{\alpha_{u}}\right)\,X\,Y$ (41)

This gives non-perturbative information about anomalies. Conversely, if the
correlators in the two regions are forbidden by unbroken symmetries, the
correlator (34) must vanish as $z\to\infty$.

That means that up to a phase [4, 5, 6] that is arbitrary and that we will set
to zero, we can use (34-37) to calculate any correlator using (32) whether or
not it is non-zero in perturbation theory and the result is

${\begin{array}[]{c}\displaystyle\Braket{0}{T\,\prod_{u=1}^{n_{o}}O_{\alpha_{u}s_{u}}(x_{u})}{0}=\left(\frac{\xi
m}{2\pi}\right)^{n_{o}}\left(\prod_{u}N_{\alpha_{u}}\right)\\\
\displaystyle\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}K_{0}\left(m_{j}\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]\end{array}}$
(42)

This makes sense because when $b^{\prime}=\emptyset$ with no $\mathcal{D}$
fields, there are no global chiral symmetries that are not broken by the gauge
boson mass terms. So there are no constraints on the use of (34-37). Because
the bosonization mass $m$ is arbitrary, it will cancel explicitly in (42), so
we can also write (42) entirely in terms of the $m_{j}$s as

${\begin{array}[]{c}\displaystyle\Braket{0}{T\,\prod_{u=1}^{n_{o}}O_{\alpha_{u}s_{u}}(x_{u})}{0}=\left(\frac{\xi}{2\pi}\right)^{n_{o}}\left(\prod_{u}M_{\alpha_{u}}\right)\\\
\displaystyle\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}K_{0}\left(m_{j}\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]\end{array}}$
(43)

where

$M_{\alpha}=\prod_{j\in b}m_{j}^{\eta_{\alpha j}^{2}}$ (44)

Equations (31), (42), (44), and (43) are the basic technical results of this
note. (31,42) and (44,43) are equivalent, but (31,42) is more convenient for
taking $m_{j}$s to zero which we will do in the next section. The $m_{j}$
dependence in (31) and (44) is useful in understanding the structure of the
general theory.

Now we can use (31,42) to analyze the general theory for
$b^{\prime}\neq\emptyset$. While we could do the general analysis directly,
constructing the general theory as the limit of (42) as some gauge couplings
go to zero automatically includes the nonperturbative effects of the gauge
anomalies and will give us insights into their structure. It may seem
surprising that this works at all because the corresponding limit in 3+1
dimensions is quite dangerous. But here everything goes smoothly because we
have properly included the important dimensional parameters in (31).

In particular, look at the limit $m_{j}\to 0$ for some $j$ in (42). Because of
the product structure of the correlator, we can focus just on the factors that
depend on $j$:

$\left(\prod_{u}\left(\frac{m_{j}}{m}\right)^{\eta_{\alpha_{u}j}^{2}}\right)\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}K_{0}\left(m_{j}\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]$
(45)

If $m_{j}\sqrt{-(x_{u}-x_{u^{\prime}})^{2}}$ is much less than $1$ for all the
coordinate pairs, we can approximate the Bessel functions by logs:

$\left(\prod_{u}\left(\frac{m_{j}}{m}\right)^{\eta_{\alpha_{u}j}^{2}}\right)\prod_{u\neq
u^{\prime}}\exp\left[2s_{u}s_{u^{\prime}}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\log\left(m_{j}\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]$
(46)

This can be written

$\left(\frac{m_{j}}{m}\right)^{\left(\sum_{u}\eta_{\alpha_{u}j}^{2}\right)+\left(\sum_{u\neq
u^{\prime}}2s_{u}s_{u^{\prime}}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\right)}\prod_{u\neq
u^{\prime}}\exp\left[2s_{u}s_{u^{\prime}}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\log\left(m\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]$
(47)
$=\left(\frac{m_{j}}{m}\right)^{\left(\sum_{u}s_{u}\eta_{\alpha_{u}j}\right)^{2}}\prod_{u\neq
u^{\prime}}\exp\left[2s_{u}s_{u^{\prime}}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\log\left(m\sqrt{-(x_{u}-x_{u^{\prime}})^{2}+i\epsilon}\right)\right]$
(48)

Now we can take the limit $m_{j}\to 0$. The limit vanishes unless

$Q_{j}=\sum_{u}s_{u}\eta_{\alpha_{u}j}=0$ (49)

This is expected from (33). In (49), $Q_{j}$ is the total $j$ chiral charge of
the product of operators in the correlator. The limit is just the statement
that the correlator vanishes if the product of operators carries an unbroken
chiral symmetry. But (48) shows us exactly how the limit is approached.

If $Q_{j}$ vanishes, the first factor in (48) is $1$, the $m_{j}$ dependence
disappears, and the bosonization mass $m$ shows up in the second factor in the
expected way for a bosonization field. Notice that it was critical here to
include the $N(\alpha)$ factor.

We can now find the correlators for a general model by starting with all gauge
couplings non-zero and then taking $g_{j}\to 0$ for $j\in b^{\prime}$, so in
general the collections of $n_{o}$ operators that have non-zero correlators
will satisfy

$Q_{j}=\sum_{u}s_{u}\eta_{\alpha_{u}j}=0\quad\forall\,j\in b^{\prime}$ (50)

The correlators are then simply

${\begin{array}[]{c}\displaystyle\Braket{0}{T\,\prod_{u=1}^{n_{o}}O_{\alpha_{u}s_{u}}(x_{u})}{0}=\left(\frac{\xi
m}{2\pi}\right)^{n_{o}}\left(\prod_{u}N_{\alpha_{u}}\right)\\\
\displaystyle\prod_{u\neq
u^{\prime}}\exp\left[-2s_{u}s_{u^{\prime}}\sum_{j}\eta_{\alpha_{u}j}\eta_{\alpha_{u^{\prime}}j}\Delta_{j}\left(x_{u}-x_{u^{\prime}}\right)\right]\end{array}}$
(51)

where as in (37)

$\Delta_{j}(x)=\left\\{{\begin{array}[]{ll}K_{0}\left(m_{j}\sqrt{-x^{2}+i\epsilon}\right)&\mbox{for
$j\in b$}\\\ -\log\left(m\sqrt{-x^{2}+i\epsilon}\right)&\mbox{for $j\in
b^{\prime}$}\end{array}}\right.$ (52)

##  3 Anomalies

Obviously, as $m_{j}\propto g_{j}\to 0$, the gauge anomalies associated with
$A^{\mu}_{j}$ disappear. But the requirement of (50) puts interesting
constraints on the form of the remaining anomalies for $m_{j}=0$. In general a
set of operators $O_{\alpha_{u}s_{u}}$ satisfying (50) can be decomposed into
irreducible sets which cannot be split into subsets satisfying (50). These
always include any single operators $\\{O_{\alpha+}\\}$ and
$\\{O_{\alpha-}\\}$ for which $\eta_{\alpha j}=0\,\forall j\in b^{\prime}$ and
the $\pm$ pairs, $\\{O_{\alpha+},O_{\alpha-}\\}$ for each $\alpha$ for which
$\eta_{\alpha j}\neq 0$ for any $j\in b^{\prime}$. I will refer to any other
irreducible sets as “anomaly sets” because they are related to the remaining
gauge anomalies. They are characterized by a set of integers ${n_{\alpha}}$:

$S({n_{\alpha}})$ consists of $|n_{\alpha}|$ copies of $O_{\alpha\mathop{\rm
sgn}(n_{\alpha})}$ for each $\alpha$ (53)

In terms of the $n_{\alpha}$, (50) is

$\sum_{\alpha}|n_{\alpha}|\mathop{\rm sgn}(n_{\alpha})\eta_{\alpha
j}=\sum_{\alpha}n_{\alpha}\eta_{\alpha j}=0\quad\forall\,j\in b^{\prime}$ (54)

Then (15) implies that $n_{\alpha}$ is a linear combination of $\eta_{\alpha
j}$ for $j\in b$,

$n_{\alpha}=\sum_{j\in b}\mathcal{N}_{j}\eta_{\alpha j}$ (55)

It makes sense that the irreducible anomaly sets depend only on the remaining
gauge couplings.

##  4 Examples

If there is only one gauge boson, there is only one $\eta_{\alpha j}$ for
$j\in b$ — call it $\eta_{\alpha 1}$. Then (55) is

$n_{\alpha}\propto\eta_{\alpha 1}$ (56)

which fixes the irreducible anomaly sets (if they exist) up to a sign (which
is just complex conjugation of all the bilinears). But $n_{\alpha}$ exists if
an only if the components of $\eta_{\alpha 1}$ are commensurate.

Here is the 2-flavor Schwinger model with $b=\\{1\\}$:

$\eta_{\alpha 1}=\left(\begin{array}[]{c}1/\sqrt{2}\\\
1/\sqrt{2}\end{array}\right)\quad\eta_{\alpha
2}=\left(\begin{array}[]{c}-1/\sqrt{2}\\\ 1/\sqrt{2}\end{array}\right)\quad
n_{\alpha}=\left(\begin{array}[]{c}\pm 1\\\ \pm 1\end{array}\right)$ (57)

The irreducible anomaly sets are $\\{O_{1\pm},O_{2\pm}\\}$.

Here is a similar model again with $b=\\{1\\}$ but gauge couplings that differ
by a factor of 2.

$\eta_{\alpha 1}=\left(\begin{array}[]{c}2/\sqrt{2}\\\
1/\sqrt{2}\end{array}\right)\quad\eta_{\alpha
2}=\left(\begin{array}[]{c}-1/\sqrt{2}\\\ 2/\sqrt{2}\end{array}\right)\quad
n_{\alpha}=\left(\begin{array}[]{c}\pm 2\\\ \pm 1\end{array}\right)$ (58)

The irreducible anomaly sets are $\\{2\times O_{1\pm},O_{2\pm}\\}$.

Again with $b=\\{1\\}$

$\eta_{\alpha 1}=\left(\begin{array}[]{c}n_{1}/\sqrt{n_{1}^{2}+n_{2}^{2}}\\\
n_{2}/\sqrt{n_{1}^{2}+n_{2}^{2}}\end{array}\right)\quad\eta_{\alpha
2}=\left(\begin{array}[]{c}-n_{2}/\sqrt{n_{1}^{2}+n_{2}^{2}}\\\
n_{1}/\sqrt{n_{1}^{2}+n_{2}^{2}}\end{array}\right)$ (59)

Now if $n_{1}$ and $n_{2}$ are relatively prime,

$n_{\alpha}=\left(\begin{array}[]{c}\pm n_{1}\\\ \pm n_{2}\end{array}\right)$
(60)

and the irreducible anomaly sets are $\\{n_{1}\times O_{1\pm},n_{2}\times
O_{2\pm}\\}$. If $n_{1}$ and $n_{2}$ have a common factor $n_{c}$ the
irreducible anomaly sets are $\\{(n_{1}/n_{c})\times
O_{1\pm},(n_{2}/n_{c})\times O_{2\pm}\\}$.

Finally for $b=\\{1\\}$, if the gauge couplings are not commensurate — as in.

$\eta_{\alpha 1}=\left(\begin{array}[]{c}\sqrt{2/3}\\\
\sqrt{1/3}\end{array}\right)\quad\eta_{\alpha
2}=\left(\begin{array}[]{c}-\sqrt{1/3}\\\ \sqrt{2/3}\end{array}\right)$ (61)

there are no anomaly sets. [7]

If there is more than gauge boson, more complicated scenarios are possible. If
the components of $\eta_{\alpha j}$ are commensurate for each $j\in b$, there
is an independent pair for anomaly sets for each $j\in b$ and they may be
reducible. For example, with two gauge bosons, $b={1,2}$, we could have a
“diagonal color” model: [8, 9, 10, 11, 12]

$\eta_{\alpha 1}=\left(\begin{array}[]{c}1/\sqrt{2}\\\ -1/\sqrt{2}\\\
0\end{array}\right)\quad\eta_{\alpha 2}=\left(\begin{array}[]{c}1/\sqrt{6}\\\
1/\sqrt{6}\\\ -2/\sqrt{6}\end{array}\right)\quad\eta_{\alpha
3}=\left(\begin{array}[]{c}1/\sqrt{3}\\\ 1/\sqrt{3}\\\
1/\sqrt{3}\end{array}\right)$ (62)

The irreducible anomaly sets are $\\{O_{1\pm},O_{2\mp}\\}$,
$\\{O_{2\pm},O_{3\mp}\\}$, and $\\{O_{3\pm},O_{1\mp}\\}$.

But incommensurate charges can further eliminate all the anomaly sets as in
(61) or just limit them, as in this contrived example (again with $b={1,2}$):

$\eta_{\alpha 1}=\left(\begin{array}[]{c}1/2\\\ \left(2+\sqrt{2}\right)/4\\\
\left(2-\sqrt{2}\right)/4\end{array}\right)\quad\eta_{\alpha
2}=\left(\begin{array}[]{c}-1/2\\\ \left(2+\sqrt{2}\right)/4\\\
\left(2-\sqrt{2}\right)/4\end{array}\right)\quad\eta_{\alpha
3}=\left(\begin{array}[]{c}1/\sqrt{2}\\\ -1/2\\\ 1/2\end{array}\right)$ (63)

Here because the only linear combination of $\eta_{\alpha 1}$ and
$\eta_{\alpha 2}$ with commensurate components is $\eta_{\alpha
1}+\eta_{\alpha 2}$, the only irreducible anomaly sets are
$\\{O_{2\pm},O_{3\pm}\\}$.

##  5 Conclusions

While the physics is trivial, the correlators in generalizations of the
massless Schwinger model depend in nontrivial ways on the gauge boson masses.
I believe that the complete analysis of the mass dependence in this note
clarifies the relationship between models with different number of gauge
bosons and chiral anomalies.

##  6 Acknowledgements

I am grateful to Igor Klebanov for helpful comments. This project has received
support from the European Union’s Horizon 2020 research and innovation
programme under the Marie Skodowska-Curie grant agreement No 860881-HIDDeN.

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