# Cuspidality criterion for symmetric powers
of automorphic representations of GL(2)
over function fields

Luis Lomelí  and  Javier Navarro

###### Abstract.

Given a cuspidal automorphic representation of GL(2) over a global function
field, we establish a comprehensive cuspidality criterion for symmetric
powers. The proof is via passage to the Galois side, possible over function
fields thanks to the Langlands correspondence of L. Lafforgue and additional
results of G. Henniart and B. Lemaire. Our work is guided by the number fields
results of Kim-Shahidi and Ramakrishnan.

###### Key words and phrases:

Langlands Correspondence; Function Fields; Symmetric Powers

## Introduction

Ever since the early beginnings of the Langlands Program, symmetric powers of
a cuspidal representation $\pi$ of ${\rm GL}_{2}(\mathbb{A}_{F})$ over a
global field $F$ with ring of adèles $\mathbb{A}_{F}$ have earned a special
place in the theory of automorphic forms and representations; Langlands
himself noted that their automorpy implies the Ramanujan Conjecture for
general linear groups, to say the least. The literature on the subject is very
general and spread over a diverse set of articles. We here have the particular
interest of establishing a complete cuspidality criterion for ${\rm
Sym}^{n}\pi$ over a global function field.

Gelbart and Jacquet established the adjoint lift of $\pi$ to ${\rm
GL}_{3}(\mathbb{A}_{F})$ in the generality of a global field in [2], namely
${\rm Ad}\,\pi={\rm Sym}^{2}\pi\otimes\omega_{\pi}^{-1}$, where $\omega_{\pi}$
is the central character of $\pi$. Then ${\rm Sym}^{2}\pi$ is automorphic and
can be written as an isobaric sum. The cuspidality criterion for the symmetric
square is already present in their work, in particular, ${\rm Sym}^{2}\pi$ is
cuspidal if and only if $\pi$ is dihedral.

In the number field case, the automorphy of ${\rm Sym}^{3}\pi$ was established
by Kim and Shahidi in [10], where it is also first noted that it is cuspidal
unless $\pi$ is dihedral or tetrahedral. They obtain further breakthroughs in
[7, 9]; Kim proving the automorphy of ${\rm Sym}^{4}\pi$, then in Kim and
Shahidi one can find a thorough cuspidality criterion including the
observation that ${\rm Sym}^{4}\pi$ is non-cuspidal if $\pi$ is dihedral,
tetrahedral or octahedral. Assuming all $\operatorname{Sym}^{n}\pi$ modular,
and still in characteristic zero, Ramakrishnan provides the criterion for
${\rm Sym}^{5}\pi$ and ${\rm Sym}^{6}\pi$ as well as noting that this is
enough for higher symmetric powers [15].

In contrast to the literature, we work with $\ell$-adic instead of complex
representations on the Galois side, and tackle the function field case. In
fact, from this point onwards, $F$ denotes a global function field of
characteristic $p$ and $\ell$ is a prime different from $p$. In this scenario,
the global Langlands correspondence is a landmark result of V. Drinfeld for
${\rm GL}_{2}$ [1] and L. Lafforgue for ${\rm GL}_{n}$ [11]. In particular, as
a consequence of their work, we know that ${\rm Sym}^{n}\pi$ is automorphic in
positive characteristic. Thanks to the available machinery over function
fields, our proofs are unconditional and we provide a comprehensive criterion
including icosahedral representations.

Our main results on the automorphic side of the Langlands correspondence are
summarized in Theorem 5.1, which provides a cuspidality criterion for the
symmetric powers. In order to phrase the criterion in a succint way, we let
$M$ be the maximal power such that ${\rm Sym}^{M}\pi$ is cuspidal, writing
$M=\infty$ in case every symmetric power of $\pi$ is cuspidal.

Automorphic Criterion. _Let $\pi$ be a cuspidal representation of ${\rm
GL}_{2}(\mathbb{A}_{F})$. Then ${\rm Sym}^{6}\pi$ is cuspidal if and only if
$M=\infty$. If ${\rm Sym}^{6}\pi$ is non-cuspidal, then $\pi$ admits the
following classification._

  * _M=1: $\pi$ is dihedral._

  * _M=2: $\pi$ is tetrahedral._

  * _M=3: $\pi$ is octahedral._

  * _M=5: $\pi$ is icosahedral._

_Additionally, ${\rm Sym}^{4}\pi$ is cuspidal if and only if ${\rm
Sym}^{5}\pi$ is as well._

However, the proof of our automorphic criterion is via passage to the Galois
side. Possible, thanks to the work of L. Lafforgue [11] as expanded to
$\ell$-adic representations by Henniart and Lemaire in [3], results that we
summarize in § 5. Given a cuspidal $\pi$ of ${\rm GL}_{2}(\mathbb{A}_{F})$
corresponding to an irreducible $2$-dimensional $\ell$-adic Galois $\sigma$
via the global Langlands correspondence

$\pi\longleftrightarrow\sigma,$

the automorphic representation ${\rm Sym}^{n}\pi$ is cuspidal if and only if
${\rm Sym}^{n}\sigma$ is irreducible. With this in mind, our main result on
the Galois side is a reducibility criterion, where $M$ in this setting denotes
the maximal irreducible symmetric power. It is Theorem 4.9, whose contents are
as follows.

Galois criterion. _Let $\sigma$ be an irreducible 2-dimensional $\ell$-adic
Galois representation, then the following are equivalent:_

  * (i)

_$\sigma$ has open kernel._

  * (ii)

_There exists an integer $n\geq 2$ such that $\,{\rm Sym}^{n}\sigma$ is
reducible._

  * (iii)

_${\rm Sym}^{6}\sigma$ is reducible._

_Specifically, we have the following classification when these properties are
met._

  * _M=1: $\sigma$ is dihedral._

  * _M=2: $\sigma$ is tetrahedral._

  * _M=3: $\sigma$ is octahedral_.

  * _M=5: $\sigma$ is icosahedral._

_Additionally, ${\rm Sym}^{5}\sigma$ is reducible if and only if ${\rm
Sym}^{4}\sigma$ is as well._

Let us make a few comments around the statement of the criterion. First, from
general representation theoretical results paired with properties of
$L$-functions, we deduce that the heart of the irreducibility criteria is
contained in the cases of $n=2,3,4$ and $6$. We next prove that if ${\rm
Sym}^{n}\sigma$ is reducible then there exists an open subgroup $H$ of $G_{F}$
such that the restriction of $\sigma$ to $H$ is reducible, say
$\sigma_{H}=\mu_{1}\oplus\mu_{2}$ for $\ell$-adic characters $\mu_{1}$ and
$\mu_{2}$. Now, from observations made by Henniart-Lemaire on $\ell$-adic
representations [3], $\mu_{1}$ and $\mu_{2}$ have open kernels. We then
conclude that $\sigma$ has open kernel. In particular, the image $J$ of
$\sigma$ in ${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$, considering
$\sigma$ as a homomorphism into ${\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})$,
is a finite group. In this case, $\sigma$ is defined to be dihedral,
tetrahedral, octahedral, or icosahedral, according to $J$ being one of these
finite groups, and observe that it cannot be abelian due to the irreducibility
of $\sigma$. We note that this classification depends only on the isomorphism
class of $\sigma$.

Let us now present the contents of the article in a more detailed fashion. We
begin with a section on the preliminaries. In particular, we recall the
representation theoretic Clebsch-Gordan formulas in § 1.2. Let us interject at
this point and mention that we additionally include an Appendix, where we
gather results on Artin $L$-functions and reducibility, and formulate a useful
property on subrepresentations. With these results in mind, we proceed in § 2
to prove two general useful lemmas concerning the reducibility of symmetric
powers. First, if $\operatorname{Sym}^{N}(\sigma)$ is reducible for some
positive integer $N$, then it is reducible hence forwards for $n\geq N$. This
tells us that the maximal irreducible symmetric power $M$, when it exists as a
positive integer, is indeed determined by the fact that
$\operatorname{Sym}^{M}(\sigma)$ is irreducible, while
$\operatorname{Sym}^{M+1}(\sigma)$ is reducible. When $M$ is finite, the
second lemma further reduces to $M<6$. However, there are representations for
which $M=\infty$, i.e., every $\operatorname{Sym}^{n}(\sigma)$ is irreducible,
cf. Igusa [5].

For $n<6$, the irredubility criteria for $\operatorname{Sym}^{n}(\sigma)$ are
examined in § 3. As we shall see by passing to the automorphic side in § 5,
our cuspidality criteria align with the well established results of Gelbart-
Jacquet over global fields and Kim-Shahidi over number fields. For general
$n$, we are very much influenced by the number fields approach of
Ramakrishnan.

Specifically, for the case of $n=2$, we show that
$\operatorname{Sym}^{2}(\sigma)$ is reducible if and only if
$\sigma\cong\sigma\otimes\chi$ for some non-trivial quadratic character
$\chi$. In this case, we find that $H=\ker\chi$ leads to $\sigma_{H}$ being
reducible. Additionally, the image of $H$ in ${\rm
PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ is an abelian index-two subgroup of
$J$, implying that $J$ must be dihedral.

Moving on to the case of $n=3$, we establish that if $\sigma$ is non-dihedral,
then $\operatorname{Sym}^{3}(\sigma)$ is reducible if and only if
$\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{2}(\sigma)\otimes\mu$
for some non-trivial cubic character $\mu$. In this situation, if
$H^{\prime}=\ker\mu$, then $\sigma_{H^{\prime}}$ is dihedral, leading to the
existence of an index-two subgroup $H$ of $H^{\prime}$ such that $\sigma_{H}$
is reducible. We observe that the image of $H^{\prime}$ in ${\rm
PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ is an index-three dihedral subgroup of
$J$, then $J$ is tetrahedral.

Finally, for $n=4$, we prove that if $\sigma$ is neither dihedral nor
tetrahedral, then $\operatorname{Sym}^{4}(\sigma)$ is reducible if and only if
there exists a non-trivial quadratic character $\chi$ such that
$\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\sigma)\otimes\chi$.
In this case, the kernel $H^{\prime}=\ker\chi$ results in
$\sigma_{H^{\prime}}$ being tetrahedral. Consequently, there exists an index-
six subgroup $H$ of $H^{\prime}$ such that $\sigma_{H}$ is reducible. In this
case, the group $J$ is necessarily octahedral. It is noteworthy that in each
of these cases, the image of $H$ in ${\rm
PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ is an abelian group appearing in a
subnormal series with abelian factors of $J$, which reflects the solvability
of the dihedral, tetrahedral and octahedral groups.

The non-solvable icosahedral case requires a separate treatment. We explore
this case in § 4 and take the opportunity to gather the remaining reducibility
Galois criteria that involve $\operatorname{Sym}^{6}(\sigma)$. There we assume
$\sigma$ is not solvable polyhedral, i.e., neither dihedral, tetrahedral nor
octahedral. We begin by proving that $\operatorname{Sym}^{6}(\sigma)$ is
reducible if and only if $\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma)\otimes\sigma^{\prime}$, for another irreducible 2-dimensional
$\ell$-adic representation $\sigma^{\prime}$. The existence of the
representation $\sigma^{\prime}$ was expected in characteristic $p$ since it
appears in the characteristic zero case in the literature, cf. [8], [15] and
[18]. We then prove, and this is crucial, that
$\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\widetilde{\sigma})$,
where $\widetilde{\sigma}=\sigma^{\prime}\otimes\xi$, for some character
$\xi$. Furthermore, for suitable bases, there exists an isomorphism
$\operatorname{Sym}^{3}(\sigma)\xrightarrow{\,\sim\,}\operatorname{Sym}^{3}(\widetilde{\sigma})$
of the form $\operatorname{Sym}^{3}(g)$, for some $g\in{\rm
GL}_{2}(\overline{\mathbb{Q}}_{\ell})$. Then, using that the homomorphism
$\operatorname{Sym}^{3}\colon{\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})\to{\rm
GL}_{4}(\overline{\mathbb{Q}}_{\ell})$ has finite kernel, we deduce that there
exists an open subgroup $H^{\prime}$ (which shall be denoted by $H$ in § 4) of
$G_{F}$ such that $\sigma_{H^{\prime}}\cong\widetilde{\sigma}_{H^{\prime}}$.
From the relation $\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma)\otimes\sigma^{\prime}$, we conclude that
$\operatorname{Sym}^{5}(\sigma_{H^{\prime}})$ is reducible, obtaining in this
way an open subgroup $H$ of $G_{F}$ such that $\sigma_{H}$ is reducible.

We conclude with a passage from irreducible Galois $\sigma$ to cuspidal
automorphic $\pi$ via the global Langlands correspondence of L. Lafforgue
[11], incorporating the results of Henniart-Lemaire [3]. We do this in § 5,
after setting up the preliminaries. This enables us to obtain the Automorphic
Criterion mentioned above, as well as more delicate cuspidality criteria for
$\operatorname{Sym}^{n}(\pi)$ corresponding in turn to the Galois treatment
for the cases when $n<6$ and the separate discussion involving the case of
$n=6$ and icosahedral representations.

### Acknowledgments

We thank Guy Henniart for mathematical communications. The first author is
grateful to the Institut des Hautes Études Scientifiques for the hospitality
provided during a summer visit in 2023, when this article was finalized. He
was supported in part by FONDECYT Grant 1212013. The second author was
partially supported by USM PIIC Initiation to Scientific Research Program.

##  1\. Preliminaries

### 1.1.

We let $F$ be a global function field of characteristic $p$ and fix a
separable closure $\overline{F}$. We denote by $\mathbb{A}_{F}$ the ring of
adèles of $F$. We also fix a prime number $\ell\not=p$ and an algebraic
closure $\overline{\mathbb{Q}}_{\ell}$ of the $\ell$-adic numbers
$\mathbb{Q}_{\ell}$. We write $G_{F}$ and $\mathcal{W}_{F}$ for the absolute
Galois group ${\rm Gal}(\overline{F}/F)$ and the Weil group of $F$
corresponding to $\overline{F}$, respectively. We assume all separable field
extensions $E/F$ lie inside $\overline{F}$.

Let $\tau$ be either a homomorphism $G_{F}\to{\rm GL}(V)$ or
$\mathcal{W}_{F}\to{\rm GL}(V)$, where $V$ is a
$\overline{\mathbb{Q}}_{\ell}$-vector space endowed with the topology induced
from $\overline{\mathbb{Q}}_{\ell}$. We say that $\tau$ is an $\ell$-adic
representation of $G_{F}$, resp. of $\mathcal{W}_{F}$, if it is continuous and
unramified at almost every place of $F$. If $H$ is a subgroup of $G_{F}$, we
use $\tau_{H}$ to denote the restriction $\tau|_{H}$. If $E/F$ is a separable
field extension, we write $\tau_{E}$ for the restriction $\tau_{G_{E}}$.

We can view a finite-dimensional $\ell$-adic representation $\tau\colon
G_{F}\to{\rm{GL}}(V)$ as a continuous homomorphism $\tau\colon
G_{F}\to{\rm{GL}}_{m}(\overline{\mathbb{Q}}_{\ell})$, where $m$ is the
dimension of $V$, after choosing a basis. For every integer $n\geq 1$, we have
the $n$-th symmetric power map

$\operatorname{Sym}^{n}\colon{\rm{GL}}_{m}(\overline{\mathbb{Q}}_{\ell})\to{\rm{GL}}_{N}(\overline{\mathbb{Q}}_{\ell}),\text{
where }N=\left(\begin{array}[]{c}m+n-1\\\ n\end{array}\right).$

It is well known, since Chevalley, that if $\tau$ is semisimple then
$\operatorname{Sym}^{n}(\tau)$ is semisimple.

Additionally, we introduce the following notation:

$A^{n}(\tau)=\operatorname{Sym}^{n}(\tau)\otimes\omega_{\tau}^{-1},\text{
where }\omega_{\tau}=\det(\tau).$

In the particular case when $m=2$, we shall write ${\rm{Ad}}(\tau)$ instead of
$A^{2}(\tau)$,

${{\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})}$${{\rm
GL}_{3}(\overline{\mathbb{Q}}_{\ell})}$${{\rm
PGL}_{2}(\overline{\mathbb{Q}}_{\ell})}$$\scriptstyle{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}{\rm
Ad}}}$

since it is the adjoint map from ${\rm GL}_{2}$ to ${\rm GL}_{3}$ of Gelbart-
Jacquet [2]. We further observe that

(1.1) $A^{1}(\tau)=\tau\otimes\omega_{\tau}^{-1}\cong\tau^{\vee},$

where $\tau^{\vee}$ denotes the contragradient representation of $\tau$.

### 1.2.

Let us recall the Clebsch-Gordan formulas for symmetric powers of
${\rm{GL}}_{2}(\overline{\mathbb{Q}}_{\ell})$. Start by taking
$\operatorname{Sym}^{1}={\rm{Id}}$. Next, if
$g\in{\rm{GL}}_{2}(\overline{\mathbb{Q}}_{\ell})$ and $i$, $m$ are non-
negative integers such that $i\leq m/2$, then

$\operatorname{Sym}^{i}(g)\otimes\operatorname{Sym}^{m-i}(g)\cong\bigoplus_{j=0}^{i}\operatorname{Sym}^{m-2j}(g)\otimes\det(g)^{j}.$

Here the isomorphism is obtained via matrix conjugation.

### 1.3.

We work with a given $2$-dimensional irreducible $\ell$-adic Galois
representation $\sigma$. When viewed as a homomorphism $\sigma\colon
G_{F}\to{\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})$, we let ${\rm proj}$ be
the canonical projection from ${\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})$ to
${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$.

${G_{F}}$${{\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})}$${{\rm
PGL}_{2}(\overline{\mathbb{Q}}_{\ell})}$$\scriptstyle{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\sigma}}$$\scriptstyle{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}{\rm
proj}}}$$\scriptstyle{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}{\rm
proj}\cdot\sigma}}$

Let $J$ be the image ${\rm proj}(\sigma(G_{F}))$. In contrast to the complex
finite-dimensional case, where a continuous homomorphism $G_{F}\to{\rm
GL}_{m}(\mathbb{C})$ always has open kernel, this may not be the case for
$\ell$-adic representations. In particular, $J$ may or may not be finite.

In the case of finite image, we define $\sigma$ to be dihedral, tetrahedral,
octahedral, or icosahedral, according to $J$ being one of these finite groups.
We note that this classification depends only on the isomorphism class of
$\sigma$. One of our main results shows that $J$ being finite is equivalent to
$\sigma$ having open kernel, and in turn equivalent to
$\operatorname{Sym}^{n}(\sigma)$ being reducible for some $n$.

We adopt a similar terminology for $\ell$-adic representations of the Weil
group. Namely, if $\sigma\colon\mathcal{W}_{F}\to{\rm
GL}_{2}(\overline{\mathbb{Q}}_{\ell})$ is an irreducible $\ell$-adic
representation such that ${\rm proj}(\sigma(\mathcal{W}_{F}))$ is finite, then
we define $\sigma$ to be dihedral, tetrahedral, octahedral, or icosahedral,
depending on ${\rm proj}(\sigma(\mathcal{W}_{F}))$. These definitions are in
accordance to the complex case found in [12]. We point out that if
$\sigma\colon\mathcal{W}_{F}\to{\rm GL}_{2}(\mathbb{C})$ is a continuous
irreducible representation then, as noted by Langlands in [_loc. cit._], the
image ${\rm proj}(\sigma(\mathcal{W}_{F}))$ is a finite subgroup of ${\rm
PGL}_{2}(\mathbb{C})$.

##  2\. Reducibility Criteria for Symmetric Powers of Galois Representations:
General Notions

In this section we address two general lemmas on the irreducibility of the
symmetric powers of an irreducible two-dimensional $\ell$-adic representation

$\sigma\colon G_{F}\to{\rm{GL}}(V).$

Their proofs involve the general representation theoretic Clebsch-Gordan
formulas recalled in § 1.2, coupled with reducibility results obtained via
properties of Artin $L$-functions summarized in Appendix A.

### 2.1.

The first of the two lemmas allows us to make precise the notion of maximal
irreducible symmetric power of $\sigma$. We write $M=\infty$ in case every
symmetric power of $\sigma$ is irreducible. Otherwise, it is the positive
integer $M$ for which ${\rm Sym}^{M}(\sigma)$ is irreducible, while ${\rm
Sym}^{M+1}(\sigma)$ is reducible.

######  Lemma 2.1.

Assume that $\operatorname{Sym}^{N}(\sigma)$ is reducible for a given integer
$N>1$. Then $\operatorname{Sym}^{n}(\sigma)$ is reducible for all $n\geq N$.

###### Proof.

We assume $\operatorname{Sym}^{n}(\sigma)$ is reducible, and we know that it
is semisimple, hence we can write
$\operatorname{Sym}^{n}(\sigma)=\sigma_{1}\oplus\sigma_{2}$, for
subrepresentations $\sigma_{1}$ and $\sigma_{2}$. By Clebsch-Gordan, we have
that

$\sigma\otimes\operatorname{Sym}^{n+1}(\sigma)=\operatorname{Sym}^{n+2}(\sigma)\oplus\operatorname{Sym}^{n}(\sigma)\otimes\omega_{\sigma}.$

Then $\sigma_{i}\otimes\omega_{\sigma}$ is a subrepresentation of
$\sigma\otimes\operatorname{Sym}^{n+1}(\sigma)$ for $i=1,2$.

If $\operatorname{Sym}^{n+1}(\sigma)$ were irreducible, then
$\operatorname{Sym}^{n+1}(\sigma^{\vee})$ would be a subrepresentation of
$\sigma\otimes\sigma_{i}^{\vee}\otimes\omega_{\sigma}^{-1}\cong(\sigma\otimes\sigma_{i})^{\vee}$
for each $i=1,2$, by Property (A.2). In particular,

$\dim\sigma\otimes\sigma_{i}=2\dim\sigma_{i}\geq\dim\operatorname{Sym}^{n+1}(\sigma)=n+2,$

for $i=1,2$, but

$\dim\sigma_{1}+\dim\sigma_{2}=\dim\operatorname{Sym}^{n}(\sigma)=n+1.$

Therefore, $\operatorname{Sym}^{n+1}(\sigma)$ must also be reducible. ∎

An interesting fact is the existence of representations with $M=\infty$ and
whose corresponding image $J$ in ${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$
is infinite. For instance, for $\ell$-adic representations associated with
certain elliptic curves defined over $F$, Igusa proved in [5] an analogous
statement to the well known Serre’s open image theorem [16].

### 2.2.

The second lemma significantly reduces the number of cases required in the
reducibility criterion.

######  Lemma 2.2.

If $M$ is finite, then $M=1,2,3$ or $5$.

###### Proof.

If $\Pi$ is an irreducible subrepresentation of
$\operatorname{Sym}^{M+1}(\sigma)$ then, by Clebsch-Gordan, it is a
subrepresentation of $\sigma\otimes\operatorname{Sym}^{M}(\sigma)$. Since
$\operatorname{Sym}^{M}(\sigma)$ is irreducible, then
$\operatorname{Sym}^{M}(\sigma^{\vee})$ is a subrepresentation of
$\sigma\otimes\Pi^{\vee}$, by (A.2). In particular, we have that

$\dim(\sigma\otimes\Pi^{\vee})=2\dim\Pi\geq M+1.$

This last inequality, together with
$\dim\operatorname{Sym}^{M+1}(\sigma)=M+2$, forces a decomposition into
irreducible representations

$\operatorname{Sym}^{M+1}(\sigma)=\Pi_{1}\oplus\Pi_{2}\text{ for }M>1,$

where

$\dim\Pi_{2}-\dim\Pi_{1}=\left\\{\begin{array}[]{rl}1&\text{if }M\text{ is
odd,}\\\ 0&\text{if }M\text{ is even.}\end{array}\right.$

If $M$ is odd, then $\dim(\sigma\otimes\Pi_{2}^{\vee})=M+3$. By the
semisimplicity of $\sigma\otimes\Pi_{2}^{\vee}$, we have that

$\sigma\otimes\Pi_{2}^{\vee}=\operatorname{Sym}^{M}(\sigma^{\vee})\oplus\tau$

for some 2-dimensional representation $\tau$. Now, since $\Pi_{2}$ is
irreducible and $\tau$ is a subrepresentation of
$\sigma\otimes\Pi_{2}^{\vee}$, then $\Pi_{2}$ is a subrepresentation of
$\sigma\otimes\tau^{\vee}$, again by (A.2). This implies in particular that

$\dim\Pi_{2}=\frac{M+3}{2}\leq 4\ \Longrightarrow\ M\leq 5.$

If $M$ is even, then
$\dim(\sigma\otimes\Pi_{1}^{\vee})=\dim(\sigma\otimes\Pi_{2}^{\vee})=M+2$. In
this case, there exists a character $\mu$ such that

$\sigma\otimes\Pi_{1}^{\vee}=\operatorname{Sym}^{M}(\sigma^{\vee})\oplus\mu.$

Applying once again (A.2), we deduce that $\sigma^{\vee}$ is a
subrepresentation of
$\Pi_{1}^{\vee}\otimes\mu^{-1}=(\Pi_{1}\otimes\mu)^{\vee}$. Since
$\Pi_{1}\otimes\mu$ is irreducible, then $\sigma\cong\Pi_{1}\otimes\mu$; in
particular, $\dim\Pi_{1}=2$ and then we necessarily have $M=2$ in this case. ∎

Notice that in the proof of Lemma 2.2 we also allow the case of $M=1$, which
is an actual possibility as we will see in Theorem 3.1. In fact, at the end of
our study we will be able to conclude the same for $n=2$, $3$ and $5$.

##  3\. Reducibility Criteria for Symmetric Powers of
Galois Representations: ${\operatorname{Sym}^{n}}(\sigma)$, $n<6$

The cases treated in this section are well established over number fields. We
are particularly influenced by [2, 14, 15] for the symmetric square, [10, 9,
7] for the symmetric cube and fourth, and [15] for the symmetric fifth. In
contrast to the literature, we work with $\ell$-adic instead of complex
representations, and tackle the case of a function field $F$ of characteristic
$p$, with $\ell\neq p$. Specifically, all of the representations that we will
encounter in this section and the next are over the field
$\overline{\mathbb{Q}}_{\ell}$.

###  3.1. Symmetric Square

In this case, the following theorem and its corollary present the reducibility
criterion.

######  Theorem 3.1.

The following are equivalent

  1. (i)

$\operatorname{Sym}^{2}(\sigma)$ is reducible.

  2. (ii)

$\sigma$ is dihedral.

  3. (iii)

$\sigma\cong\sigma\otimes\chi$ for some non-trivial character $\chi$.

In this case, $\sigma$ has open kernel.

###### Proof.

Let us suppose first that $\sigma$ is dihedral, meaning that ${\rm
proj}(\sigma(G_{F}))$ is finite and dihedral. Since the dihedral groups have
an index two subgroup which is cyclic, we have that $\sigma(H)$ is cyclic
modulo the center of $\sigma(G_{F})$, for some index two subgroup $H$ of
$G_{F}$. In particular, $\sigma(H)$ is abelian. By Schur’s lemma, $\sigma_{H}$
is reducible. Note that $\sigma$ being irreducible, the representation
$\sigma_{H}$ is semisimple. Actually, if $c\in G_{F}\setminus H$, then
$\sigma_{H}\cong\mu\oplus\mu^{c}$, for some character $\mu$ of $H$, where

$\mu^{c}(h):=\mu(chc^{-1}),\quad h\in H.$

Let $E/F$ be the quadratic extension associated to $H$, so that $H=G_{E}$. We
observe that $\sigma$ being unramified almost everywhere, the character $\mu$
is unramified at almost every place of $E$. Thus, $\mu$ is an $\ell$-adic
character of $G_{E}$. By Lemme IV.2.7 of [3], $\mu$ has open kernel. Since
$\sigma_{E}\cong\mu\oplus\mu^{c}$ and $E/F$ is finite, we have that $\sigma$
has open kernel. Therefore, $\sigma$ is actually a complex representation.

Now, by Frobenius reciprocity

${\rm Hom}_{G_{E}}(\sigma_{E},\mu)={\rm Hom}_{G_{F}}(\sigma,{\rm
Ind}_{E}^{F}\mu).$

Since $\sigma$ is irreducible, the above being not zero implies that
$\sigma\cong{\rm Ind}_{E}^{F}\mu$. If $\chi$ is the only non-trivial character
of $G_{F}$ which is trivial on $H$, then

$\sigma\otimes\chi\cong{\rm{Ind}}_{E}^{F}\mu\otimes\chi\cong{\rm{Ind}}_{E}^{F}(\mu\otimes\chi_{E})={\rm{Ind}}_{E}^{F}(\mu)\cong\sigma,$

and (iii) is satisfied. Now suppose that (iii) holds. By comparing
determinants, we have that $\chi^{2}=1$. Then the kernel of $\chi$ is $G_{E}$,
for some quadratic extension $E/F$. If
$\varphi\colon\sigma\to\sigma\otimes\chi$ is an isomorphism, then $\varphi$ is
not a scalar, since $\chi$ is not trivial. Thus, by taking restriction to
$G_{E}$ we get a non-scalar isomorphism

$\varphi_{E}\colon\sigma_{E}\longrightarrow\sigma_{E}.$

By Schur’s lemma, we have that $\sigma_{E}$ is reducible. As above, we have
that $\sigma_{E}\cong\mu\oplus\mu^{c}$ for some character $\mu$ of $G_{E}$,
where $c$ is the non-trivial element of $G_{F}/G_{E}\cong{\rm{Gal}}(E/F)$.
Again this implies that $\sigma$ is complex and, in particular, ${\rm
proj}(\sigma(G_{F}))$ is finite. That $\sigma$ is dihedral follows then from
the fact that the dihedral groups are the only finite non-abelian subgroups of
${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ with index two abelian
subgroups.

Now let us prove that (the equivalent) (ii) and (iii) imply (i). Let $E/F$ be
a quadratic extension and $\mu$ a character of $G_{E}$ such that
$\sigma={\rm{Ind}}_{E}^{F}\mu$. In this case, for a suitable basis of $V$,
$\sigma$ is as follows

(3.1) $\sigma\colon
g\longmapsto\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}\mu(g)&0\\\
0&\mu(cgc^{-1})\end{array}\right)&\mbox{if }g\in G_{E},\\\ &\\\
\left(\begin{array}[]{cc}0&\mu(cg)\\\
\mu(gc^{-1})&0\end{array}\right)&\mbox{if }g\not\in G_{E}.\end{array}\right.$

From which we have that

$\operatorname{Sym}^{2}(\sigma)\colon
g\longmapsto\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}\mu^{2}(g)&0&0\\\
0&\mu(g)\mu(cgc^{-1})&0\\\ 0&0&\mu^{2}(cgc^{-1})\end{array}\right)&\mbox{if
}g\in G_{E},\\\ &\\\ \left(\begin{array}[]{ccc}0&0&\mu^{2}(cg)\\\
0&\mu(cg)\mu(gc^{-1})&0\\\ \mu^{2}(gc^{-1})&0&0\end{array}\right)&\mbox{if
}g\not\in G_{E}.\end{array}\right.$

From the above we readily see that $\operatorname{Sym}^{2}(\sigma)$ is
reducible. Actually, it has the one-dimensional subrepresentation given by

$\eta\colon
g\longmapsto\left\\{\begin{array}[]{ll}\mu(g)\mu(cgc^{-1})=\det(\sigma(g))&\mbox{if
}g\in G_{E},\\\ &\\\ \mu(cg)\mu(gc^{-1})=-\det(\sigma(g))&\mbox{if }g\not\in
G_{E}.\end{array}\right.$

We observe that $\eta=\omega_{\sigma}\chi$, where $\omega_{\sigma}=\det\circ\
\sigma$ and $\chi$ is the quadratic character associated to $E$. Here we
obtain

(3.2)
$\operatorname{Sym}^{2}(\sigma)\cong\omega_{\sigma}\chi\oplus{\rm{Ind}}_{E}^{F}\mu^{2}.$

Finally, let us suppose that (i) holds. Since $\operatorname{Sym}^{2}(\sigma)$
is semisimple and 3 -dimensional, it must have a one-dimensional summand, say
$\eta$. By Clebsch-Gordan

(3.3)
$\sigma\otimes\sigma=\operatorname{Sym}^{2}(\sigma)\oplus\omega_{\sigma},$

we have that $\eta$ is a subrepresentation of $\sigma\otimes\sigma$. By (A.2),
this implies that $\sigma^{\vee}$ is a subrepresentation of
$\sigma\otimes\eta^{-1}$. Then, by irreducibility, we get

$\sigma\otimes\eta^{-1}\cong\sigma^{\vee}\cong\sigma\otimes\omega_{\sigma}^{-1},$

from which we infer that $\sigma\cong\sigma\otimes\eta\omega_{\sigma}^{-1}$.
We observe that $\eta\not=\omega_{\sigma}$, since otherwise, from (3.3) the
$L$-function
$L(s,\sigma\otimes\sigma\otimes\eta^{-1})=L(s,\sigma\otimes\sigma^{\vee})$
would have a non-simple pole at $s=1$. The latter is not possible since
$\sigma$ is irreducible, see (A.1). Thus, by letting
$\chi=\eta\omega_{\sigma}^{-1}$, we have that (iii) holds.

∎

The following result is a general observation, beginning at $n=2$ because of
Theorem 3.1 and is valid thereafter due to Lemma 2.1. In terms of the maximal
irreducible symmetric power, this happens exactly when $M=1$.

######  Corollary 3.2.

The representation $\sigma$ is dihedral if and only if
$\operatorname{Sym}^{n}(\sigma)$ is reducible for $n\geq 2$.

###  3.2. Symmetric Cube

Given that $\operatorname{Sym}^{3}(\sigma)$ is reducible when $\sigma$ is
dihedral by Corollary 3.2, we complete the reducibility criterion in this case
by restricting ourselves to non-dihedral representations in the following
theorem.

######  Theorem 3.3.

Suppose that $\sigma$ is non-dihedral. Then $\operatorname{Sym}^{3}(\sigma)$
is reducible if and only if there exists some non-trivial character $\mu$ such
that

$\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{2}(\sigma)\otimes\mu;$

the condition being equivalent to $\sigma$ being tetrahedral. When this is the
case $\sigma$ has open kernel and we have the decomposition

$A^{3}(\sigma)\cong\sigma\otimes\mu\oplus\sigma\otimes\mu^{2}.$

###### Proof.

Let us suppose first that
$\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{2}(\sigma)\otimes\mu$
for some non-trivial character $\mu$. By Clebsch-Gordan

(3.4)
$\sigma\otimes\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{3}(\sigma)\oplus\sigma\otimes\omega_{\sigma}.$

Now,
$\sigma\otimes\operatorname{Sym}^{2}(\sigma)\cong\sigma\otimes\operatorname{Sym}^{2}(\sigma)\otimes\mu$
by hypothesis, so the right-hand side of (3.4) is equivalent to its twist by
$\mu$, then

$\operatorname{Sym}^{3}(\sigma)\oplus\sigma\otimes\omega_{\sigma}\cong(\operatorname{Sym}^{3}(\sigma)\oplus\sigma\otimes\omega_{\sigma})\otimes\mu\cong\operatorname{Sym}^{3}(\sigma)\otimes\mu\oplus\sigma\otimes\omega_{\sigma}\mu.$

Since $\sigma$ is non-dihedral and $\mu$ is non-trivial, by Theorem 3.1 we
have that
$\sigma\otimes\omega_{\sigma}\not\cong\sigma\otimes\omega_{\sigma}\mu$. Thus,
$\sigma\otimes\omega_{\sigma}\mu$ must be a factor of
$\operatorname{Sym}^{3}(\sigma)$. Hence, $\operatorname{Sym}^{3}(\sigma)$ is
reducible.

Now let us assume that $\operatorname{Sym}^{3}(\sigma)$ is reducible. Let
$\tau$ be a subrepresentation of $\operatorname{Sym}^{3}(\sigma)$. By (3.4),
$\tau$ is a subrepresentation of
$\sigma\otimes\operatorname{Sym}^{2}(\sigma)$. Since
$\operatorname{Sym}^{2}(\sigma)$ is irreducible, then it is a
subrepresentation of $\sigma^{\vee}\otimes\tau$. In particular, $\tau$ is not
one-dimensional. Thus, we have that

$\operatorname{Sym}^{3}(\sigma)=\tau_{1}\oplus\tau_{2},$

for some irreducbile two-dimensional representations $\tau_{1}$ and
$\tau_{2}$. Since $\sigma^{\vee}\otimes\tau_{1}$ is semisimple 4-dimensional,
and $\operatorname{Sym}^{2}(\sigma)$ is irreducible 3-dimensional, then there
exists a character $\eta_{1}$ such that

(3.5)
$\sigma^{\vee}\otimes\tau_{1}=\operatorname{Sym}^{2}(\sigma)\oplus\eta_{1}.$

This implies that $\sigma$ is a subrepresentation of
$\tau_{1}\otimes\eta_{1}^{-1}$ and then
$\sigma\cong\tau_{1}\otimes\eta_{1}^{-1}$ by irreducibility, i.e.,
$\tau_{1}\cong\sigma\otimes\eta_{1}$.

Similarly, we have that $\tau_{2}\cong\sigma\otimes\eta_{2}$ for some
character $\eta_{2}$. We obtain then

(3.6)
$\operatorname{Sym}^{3}(\sigma)\cong\sigma\otimes\eta_{1}\oplus\sigma\otimes\eta_{2}.$

Now, by the identity $\tau_{1}\cong\sigma\otimes\eta_{1}$ we get

$\sigma^{\vee}\otimes\tau_{1}\cong\sigma^{\vee}\otimes\sigma\otimes\eta_{1}\cong\operatorname{Sym}^{2}(\sigma)\otimes\eta_{1}\omega_{\sigma}^{-1}\oplus\eta_{1}.$

Thus, by (3.5) we obtain

$\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{2}(\sigma)\otimes\eta_{1}\omega_{\sigma}^{-1}.$

We observe that neither $\eta_{1}$ nor $\eta_{2}$ can be equal to
$\omega_{\sigma}$, since the pole at $s=1$ of
$L(s,\sigma\otimes(\sigma\otimes\omega_{\sigma})^{\vee}\otimes\operatorname{Sym}^{2}(\sigma))$
is simple, by dimension reasons. Also, $\eta_{1}\not=\eta_{2}$, since the pole
at $s=1$ of
$L(s,\sigma\otimes\tau_{i}^{\vee}\otimes\operatorname{Sym}^{2}(\sigma))$ is
simple, for $i=1,2$. Thus, if we let $\mu_{i}=\eta_{i}\omega_{\sigma}^{-1}$
for $i=1,2$, then $\mu_{1}\not=\mu_{2}$ are non-trivial characters such that

(3.7)
$\operatorname{Sym}^{2}(\sigma)\cong\operatorname{Sym}^{2}(\sigma)\otimes\mu_{1}\cong\operatorname{Sym}^{2}(\sigma)\otimes\mu_{2}$

and

(3.8) $A^{3}(\sigma)\cong\sigma\otimes\mu_{1}\oplus\sigma\otimes\mu_{2}.$

The first part of the theorem is proved, since $\mu_{1}$ and $\mu_{2}$ are not
trivial.

We observe that (3.7) implies in particular that $\mu_{1}$ has order three. If
$E/F$ is the cubic extension which corresponds to $\mu_{1}$, then similarly as
in § 3.1 above, (3.7) implies that $\operatorname{Sym}^{2}(\sigma_{E})$ is
reducible. Note that since $E/F$ is cyclic of order three and $\sigma$ is two-
dimensional irreducible, then $\sigma_{E}$ is irreducible. By Theorem 3.1,
$\sigma_{E}$ is dihedral. In particular, ${\rm proj}(\sigma(G_{F}))$ is
finite. It must be tetrahedral, since $A_{4}$ is the unique non-abelian finite
subgroup of ${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ with a dihedral
index three subgroup.

Now, by taking contragradient of both sides of (3.8), we have that

$A^{3}(\sigma)\otimes\omega_{\sigma}^{-1}\cong\sigma\otimes\omega_{\sigma}^{-1}\mu_{1}^{-1}\oplus\sigma\otimes\omega_{\sigma}^{-1}\mu_{2}^{-1},$

i.e.,

$A^{3}(\sigma)\cong\sigma\otimes\mu_{1}\oplus\sigma\otimes\mu_{2}\cong\sigma\otimes\mu_{1}^{-1}\oplus\sigma\otimes\mu_{2}^{-1}.$

If $\sigma\otimes\mu_{1}\cong\sigma\otimes\mu_{1}^{-1}$, then
$\sigma\otimes\mu_{1}^{2}\cong\sigma$, which is not possible since
$\mu_{1}^{2}$ is non-trivial and $\sigma$ is non-dihedral. Then, necessarily
$\sigma\otimes\mu_{1}\cong\sigma\otimes\mu_{2}^{-1}$. The latter implies that
$\mu_{1}=\mu_{2}^{-1}$, since $\sigma$ is non-dihedral. If we take, for
instance, $\mu=\mu_{2}$, then $A^{2}(\sigma)\cong A^{2}(\sigma)\otimes\mu$
and, since $\mu_{1}=\mu_{2}^{-1}=\mu_{2}^{2}$, (3.8) becomes

$A^{3}(\sigma)\cong\sigma\otimes\mu\oplus\sigma\otimes\mu^{2}.$

Theorem 3.3 is now completed. ∎

Similar to Corollary 3.2, the following result is a consequence of Theorem 3.3
and Lemma 2.1, together with the definition of $M$.

######  Corollary 3.4.

The representation $\sigma$ is tetrahedral if and only if $M=2$.

###  3.3. Fourth and Fifth Symmetric Powers

If $\operatorname{Sym}^{3}(\sigma)$ is reducible then, by Lemma 2.1, so is
$\operatorname{Sym}^{4}(\sigma)$. To complete the criterion it suffices to
inspect the case when the symmetric cube is irreducible in the next theorem.

######  Theorem 3.5.

Assume that $\operatorname{Sym}^{3}(\sigma)$ is irreducible. Then
$\operatorname{Sym}^{4}(\sigma)$ is reducible if and only if there exists a
non-trivial quadratic character $\chi$ such that

$\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\sigma)\otimes\chi;$

the condition being equivalent to $\sigma$ being octahedral. When this is the
case $\sigma$ has open kernel and

$A^{4}(\sigma)\cong{\rm{Ind}}_{E}^{F}(\omega_{\sigma_{E}}\mu)\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi,$

where $E/F$ is the quadratic extension corresponding to $\chi$ by class field
theory and $\mu$ is some cubic character of $G_{E}$.

###### Proof.

First, if
$\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\sigma)\otimes\chi$,
then clearly $A^{3}(\sigma)\cong A^{3}(\sigma)\otimes\chi$. By Clebsch-Gordan

(3.9) $\sigma\otimes A^{3}(\sigma)\cong
A^{4}(\sigma)\oplus\operatorname{Sym}^{2}(\sigma).$

Then we have

$A^{4}(\sigma)\oplus\operatorname{Sym}^{2}(\sigma)\cong
A^{4}(\sigma)\otimes\chi\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi.$

Since
$\operatorname{Sym}^{2}(\sigma)\otimes\chi\not\cong\operatorname{Sym}^{2}(\sigma)$,
we have that $\operatorname{Sym}^{2}(\sigma)\otimes\chi$ is a factor of
$A^{4}(\sigma)$. In particular, $\operatorname{Sym}^{4}(\sigma)$ is reducible.

Let us suppose now that $\operatorname{Sym}^{4}(\sigma)$, and therefore
$A^{4}(\sigma)$, is reducible. Since $A^{3}(\sigma)$ is assumed to be
irreducible, then for every subrepresentation $\tau$ of $\sigma\otimes
A^{3}(\sigma)$, $A^{3}(\sigma)$ is a subrepresentation of
$\sigma^{\vee}\otimes\tau$. In particular, $\sigma\otimes A^{3}(\sigma)$
cannot have one-dimensional subrepresentations. Thus, from (3.9) we get that

(3.10) $A^{4}(\sigma)\cong\tau\oplus\Pi,$

for some irreducible representations $\tau$ and $\Pi$ of dimensions 2 and 3,
respectively. By comparing dimensions, we deduce that there exists a two-
dimensional (necessarily irreducible) representation $\tau_{1}$ such that

$\sigma^{\vee}\otimes\Pi\cong A^{3}(\sigma)\oplus\tau_{1}.$

By (A.2), we have that $\Pi$ is a subrepresentation of
$\sigma\otimes\tau_{1}$. By comparing dimensions once again, we see that there
exists a character $\eta$ such that

(3.11) $\sigma\otimes\tau_{1}\cong\Pi\oplus\eta.$

Therefore, we must have that $\tau_{1}\otimes\eta^{-1}\cong\sigma^{\vee}$,
i.e., $\tau_{1}\cong\sigma\otimes\chi$, where $\chi=\eta\omega_{\sigma}^{-1}$.

Now, we have

$\sigma\otimes\tau_{1}\cong\sigma\otimes(\sigma\otimes\chi)\cong\operatorname{Sym}^{2}(\sigma)\otimes\chi\oplus\omega_{\sigma}\chi.$

Thus, by (3.11) we conclude that

(3.12) $\Pi\cong\operatorname{Sym}^{2}(\sigma)\otimes\chi.$

Then, by (3.9), (3.10) and (3.12) above, we obtain

(3.13) $\sigma\otimes
A^{3}(\sigma)\cong\tau\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi\oplus\operatorname{Sym}^{2}(\sigma).$

We see from (3.13) that $\chi$ is not trivial, since we are assuming that
$A^{3}(\sigma)$ is irreducible and then the pole at $s=1$ of

$L(s,\sigma\otimes
A^{3}(\sigma)\otimes\operatorname{Sym}^{2}(\sigma)^{\vee})=L(s,(\sigma\otimes\operatorname{Sym}^{2}(\sigma)^{\vee})\otimes
A^{3}(\sigma))$

must be simple, by dimension reasons. By taking contragradient of both sides
of (3.13), we have that

$\sigma\otimes
A^{3}(\sigma)\otimes\omega_{\sigma}^{-2}\cong\tau\otimes\omega_{\tau}^{-1}\oplus\operatorname{Sym}^{2}(\sigma)\otimes\omega_{\sigma}^{-2}\chi^{-1}\oplus\operatorname{Sym}^{2}(\sigma)\otimes\omega_{\sigma}^{-2}.$

Hence, we obtain

$\tau\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi\oplus\operatorname{Sym}^{2}(\sigma)\cong\tau\otimes\omega_{\tau}^{-1}\omega_{\sigma}^{2}\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi^{-1}\oplus\operatorname{Sym}^{2}(\sigma).$

In particular, we have that
$\operatorname{Sym}^{2}(\sigma)\otimes\chi\cong\operatorname{Sym}^{2}(\sigma)\otimes\chi^{-1}$.
This implies that $\chi$ is a non-trivial quadratic character by Theorem 3.3,
since we are assuming that $A^{3}(\sigma)$ is irreducible.

Let $E/F$ be the quadratic extension obtained from $\chi$ via class field
theory. By restricting to $G_{E}$, we have from (3.13) that

(3.14) $\sigma_{E}\otimes
A^{3}(\sigma_{E})\cong\tau_{E}\oplus\operatorname{Sym}^{2}(\sigma_{E})\oplus\operatorname{Sym}(\sigma_{E}).$

This implies that $A^{3}(\sigma_{E})$ is reducible. In fact, as we have
observed, otherwise the function $L(s,\sigma_{E}\otimes
A^{3}(\sigma_{E})\otimes\operatorname{Sym}^{2}(\sigma_{E})^{\vee})$ would have
pole at $s=1$ at most simple.

The representation $\sigma$ being non-dihedral, $\sigma_{E}$ is irreducible.
Thus, by §§3.1-3.2, we have that $\sigma_{E}$ has open kernel. Since $E/F$ is
finite, $\sigma$ has open kernel. Now, the representation $\sigma_{E}$ cannot
be dihedral, since none of the groups $A_{4}$, $S_{4}$ and $A_{5}$ has an
index-two subgroup which is dihedral. Actually, we necessarily have that
$\sigma$ is octahedral and $\sigma_{E}$ is tetrahedral. We then can apply
Theorem 3.3. In particular, we have that

(3.15)
$A^{3}(\sigma_{E})\cong\sigma_{E}\otimes\mu\oplus\sigma_{E}\otimes\mu^{2},$

for some character $\mu$ of $G_{E}$. Let $s$ be the non-trivial element of
$G_{F}/G_{E}\cong{\rm{Gal}}(E/F)$. Clearly, $s$ acts trivially on
$A^{3}(\sigma_{E})$, and cannot act trivially on its factors
$\sigma_{E}\otimes\mu$ and $\sigma_{E}\otimes\mu^{2}$, since in this case
$A^{3}(\sigma)$ would be reducible. Then, we have that
$(\sigma_{E}\otimes\mu)^{s}=\sigma_{E}\otimes\mu^{2}$, i.e.,
$\mu^{s}=\mu^{2}$. This implies that
$A^{3}(\sigma)\cong{{\rm{Ind}}}_{E}^{F}(\sigma_{E}\otimes\mu)$. Since $E/F$ is
the extension defined by $\chi$, we obtain

$A^{3}(\sigma)\cong A^{3}(\sigma)\otimes\chi.$

Now, from (3.15) and the fact that
$\operatorname{Sym}^{2}(\sigma_{E})\cong\operatorname{Sym}^{2}(\sigma_{E})\otimes\mu$
by Theorem 3.3 we have that

$\begin{split}\sigma_{E}\otimes
A^{3}(\sigma_{E})&\cong\left(\sigma_{E}\otimes\sigma_{E}\otimes\mu\right)\oplus\left(\sigma_{E}\otimes\sigma_{E}\otimes\mu^{2}\right)\\\
&\cong\left(\operatorname{Sym}^{2}(\sigma_{E})\oplus\omega_{\sigma_{E}}\mu\right)\oplus\left(\operatorname{Sym}^{2}(\sigma_{E})\oplus\omega_{\sigma_{E}}\mu^{2}\right).\end{split}$

By replacing the above in (3.14), we have

$\tau_{E}\cong\omega_{\sigma_{E}}\mu\oplus\omega_{\sigma_{E}}\mu^{2}.$

Therefore, $\tau\cong{{\rm{Ind}}}_{E}^{F}(\omega_{\sigma_{E}}\mu)$ and we
obtain the decomposition

$A^{4}(\sigma)\cong{{\rm{Ind}}}_{E}^{F}(\omega_{\sigma_{E}}\mu)\oplus\operatorname{Sym}^{2}(\sigma)\otimes\chi.$

We are done. ∎

Analogous to the dihedral and tetrahedral cases, Corollaries 3.2 and 3.4
respectively, for octahedral representations we have the following result.

######  Corollary 3.6.

The representation $\sigma$ is octahedral if and only if $M=3$.

Given our results thus far, the following criterion for the symmetric fifth
power is quickly obtained.

######  Theorem 3.7.

The following are equivalent:

  * (i)

$\operatorname{Sym}^{5}(\sigma)$ is reducible.

  * (ii)

$\operatorname{Sym}^{4}(\sigma)$ is reducible.

  * (iii)

$\sigma$ is either dihedral, tetrahedral or octahedral.

###### Proof.

The equivalence of (i) and (ii) follows from Lemmas 2.1 and 2.2. And, from our
prior results on $\operatorname{Sym}^{n}(\sigma)$ for $n=4,3,2$, Theorems 3.5,
3.3 and 3.1 respectively, we can infer that $\operatorname{Sym}^{4}(\sigma)$
is reducible if and only if $\sigma$ is either dihedral, tetrahedral or
octahedral. ∎

##  4\. Reducibility Criteria for Symmetric Powers of
Galois Representations: The Icosahedral Case

The main aim of this section is to establish the remaining reducibility
criteria involving the symmetric sixth, proving that if
$\operatorname{Sym}^{6}(\sigma)$ is reducible then $\sigma$ has open kernel.
Of particular interest are icosahedral represenations. We keep the notation of
§§ 2 and 3, where in particular all of our representations are $\ell$-adic.

### 4.1.

In analogy to the previous cases, arguing with Corollaries 3.2, 3.4, and now
including Corollary 3.6, it is enough to study the reducibility of
$\operatorname{Sym}^{6}(\sigma)$ when $\operatorname{Sym}^{4}(\sigma)$ is
irreducible.

######  Theorem 4.1.

If $\operatorname{Sym}^{4}(\sigma)$ is irreducible, then the following are
equivalent.

  1. (i)

$\operatorname{Sym}^{6}(\sigma)$ is reducible.

  2. (ii)

There exists an irreducible two-dimensional representation $\sigma^{\prime}$
of $G_{F}$ such that
$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}$.

The representation $\sigma^{\prime}$ is uniquely determined up to isomorphism
by (ii).

###### Proof.

Let us suppose first that
$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}$
for some $\sigma^{\prime}$. Then

(4.1)
$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\sigma\otimes{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}\cong\left(A^{3}(\sigma)\otimes\sigma^{\prime}\right)\oplus\left(\sigma\otimes\sigma^{\prime}\right).$

On the other hand, we have the identity

(4.2)
$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\operatorname{Sym}^{6}(\sigma)\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}.$

By comparing the decompositions (4.1) and (4.2), and noting that
$\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}$ is irreducible, we see
that $\operatorname{Sym}^{6}(\sigma)$ is reducible.

Let us suppose now that $\operatorname{Sym}^{6}(\sigma)$ is reducible. Let
$\tau$ be an irreducible component of $\operatorname{Sym}^{6}(\sigma)$ of
dimension $r\leq 3$, which exists by the semisimplicity of
$\operatorname{Sym}^{6}(\sigma)$. By (4.2), $\tau$ is a subrepresentation of
$\sigma\otimes\operatorname{Sym}^{5}(\sigma)$ and then
$\operatorname{Sym}^{5}(\sigma)$ is a subrepresentation of
$\sigma^{\vee}\otimes\tau$, by (A.2). Since $\operatorname{Sym}^{5}(\sigma)$
is $6$-dimensional, we necessarily have that $r=3$ and

(4.3) $\operatorname{Sym}^{5}(\sigma)\cong\sigma^{\vee}\otimes\tau.$

We observe that $\operatorname{Sym}^{6}(\sigma)\cong\Pi\oplus\tau$ for some
irreducible 4-dimensional representation $\Pi$, since
$\operatorname{Sym}^{6}(\sigma)$ cannot have irreducible summands of dimension
less than 3 by the above. Now, $\operatorname{Sym}^{5}(\sigma)$ is a
subrepresentation of $\sigma^{\vee}\otimes\Pi$, and by comparing dimensions we
deduce that there exists an irreducible 2-dimensional representation
$\sigma^{\prime}$ such that

(4.4)
$\sigma^{\vee}\otimes\Pi\cong\operatorname{Sym}^{5}(\sigma)\oplus\sigma^{\prime}.$

Using (A.2) again and comparing dimensions, we obtain that

$\Pi\cong\sigma\otimes\sigma^{\prime}.$

Then

$\sigma^{\vee}\otimes\Pi\cong\sigma^{\vee}\otimes(\sigma\otimes\sigma^{\prime})\cong\left({\rm{Ad}}(\sigma)\otimes\sigma^{\prime}\right)\oplus\sigma^{\prime}.$

By comparing with (4.4) we have that
$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}$,
i.e., (ii) holds.

Finally, let us see that the representation $\sigma^{\prime}$ is uniquely
determined by (ii). Let $\sigma^{\prime\prime}$ be another (necessarily
irreducible) representation such that $\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma)\otimes\sigma^{\prime\prime}$. Then

$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\left(A^{3}(\sigma)\otimes\sigma^{\prime}\right)\oplus\left(\sigma\otimes\sigma^{\prime}\right)\cong\left(A^{3}(\sigma)\otimes\sigma^{\prime\prime}\right)\oplus\left(\sigma\otimes\sigma^{\prime\prime}\right).$

The above being in turn isomorphic to

$\operatorname{Sym}^{6}(\sigma)\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}\cong\Pi\oplus\tau\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma},$

by comparing dimensions, we obtain that
$\Pi\cong\sigma\otimes\sigma^{\prime\prime}$. Thus, $\sigma^{\prime\prime}$ is
a subrepresentation of $\sigma^{\vee}\otimes\Pi$. By (4.4), we see that
$\sigma^{\prime\prime}\cong\sigma^{\prime}$, as wanted. ∎

######  Proposition 4.2.

Assume that $\operatorname{Sym}^{4}(\sigma)$ is irreducible and
$\operatorname{Sym}^{6}(\sigma)$ is reducible. Let $\sigma^{\prime}$ be the
representation of Theorem 4.1. Then there exists a character $\chi$ such that

$\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma^{\prime})\otimes\sigma\otimes\chi.$

In this case, we have the decomposition

$\operatorname{Sym}^{6}(\sigma)\cong\sigma\otimes\sigma^{\prime}\oplus{\rm{Ad}}(\sigma^{\prime})\otimes\chi\omega_{\sigma}.$

###### Proof.

From (4.3) we obtain

$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\sigma\otimes\sigma^{\vee}\otimes\tau\cong{\rm{Ad}}(\sigma)\otimes\tau\oplus\tau.$

On the other hand, we have that

$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\operatorname{Sym}^{6}(\sigma)\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}\cong\sigma\otimes\sigma^{\prime}\oplus\tau\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}.$

Thus,

${\rm{Ad}}(\sigma)\otimes\tau\cong\sigma\otimes\sigma^{\prime}\oplus\operatorname{Sym}^{4}(\sigma)\otimes\omega_{\sigma}.$

Therefore, $\sigma\otimes\sigma^{\prime}$ is a subrepresentation of ${\rm
Ad}(\sigma)\otimes\tau$. By (A.2), noting that ${\rm Ad}(\sigma)$ is selfdual,
we have that $\tau$ is a subrepresentation of

${\rm
Ad}(\sigma)\otimes\sigma\otimes\sigma^{\prime}\cong\left(A^{3}(\sigma)\otimes\sigma^{\prime}\right)\oplus\left(\sigma\otimes\sigma^{\prime}\right).$

We note that $\tau$ is not a subrepresentation of
$\sigma\otimes\sigma^{\prime}$, since $\sigma\otimes\sigma^{\prime}$ is
irreducible. Thus, we have that $\tau$ is a subrepresentation of
$A^{3}(\sigma)\otimes\sigma^{\prime}$. Then $A^{3}(\sigma)$ is a
subrepresentation of $\sigma^{\prime\vee}\otimes\tau$, and there exists a
(necessarily irreducible) two-dimensional representation $\tau_{0}$ such that

$\sigma^{\prime\vee}\otimes\tau\cong A^{3}(\sigma)\oplus\tau_{0}.$

Then $\tau$ is a subrepresentation of $\sigma^{\prime}\otimes\tau_{0}$ and, by
comparing dimensions, we see that $\sigma^{\prime}\otimes\tau_{0}$ contains a
character. Then, $\tau_{0}\cong\sigma^{\prime}\otimes\mu$, for some character
$\mu$, and

$\sigma^{\prime}\otimes\tau_{0}\cong\operatorname{Sym}^{2}(\sigma^{\prime})\otimes\mu\oplus\omega_{\sigma^{\prime}}\mu.$

The above implies that
$\tau\cong\operatorname{Sym}^{2}(\sigma^{\prime})\otimes\mu$. We then have
that

$\operatorname{Sym}^{5}(\sigma)\cong\sigma^{\vee}\otimes\tau\cong{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\chi,$

where $\chi=\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1}\mu$. As for the
decomposition, we note that

$\operatorname{Sym}^{6}(\sigma)\cong\Pi\oplus\tau\cong\sigma\otimes\sigma^{\prime}\oplus\operatorname{Sym}^{2}(\sigma^{\prime})\otimes\mu\cong\sigma\otimes\sigma^{\prime}\oplus{\rm{Ad}}(\sigma^{\prime})\otimes\chi\omega_{\sigma}.$

∎

###  4.2. Remarks

Over number fields, Ramakrishnan refers to the representations satisfying the
equivalente conditions of Theorem 4.1 as quasi-icosahedral, see [15]. Over
function fields, we shall make a refinement to the irreducibility criterion
presented in Theorem 4.1. We prove that these representations are icosahedral
in Theorem 4.6, by proving they have open kernel.

Furthermore, observe that the representation $\sigma^{\prime}$ of Theorem 4.1
is very peculiar. For instance, $\sigma^{\prime}\not\cong\sigma\otimes\mu$ for
every character $\mu$. Also, since
$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\chi$
is irreducible, then
${\rm{Ad}}(\sigma^{\prime})\not\cong{\rm{Ad}}(\sigma)\otimes\mu$ for every
character $\mu$. However, we shall prove in Proposition 4.3 below that
$\operatorname{Sym}^{3}(\sigma^{\prime})\cong\operatorname{Sym}^{3}(\sigma)\otimes\mu$
for some character $\mu$. This identity brings to light the fact that the
properties of the character $\chi$ of Proposition 4.2 determine it uniquely,
see Corollary 4.4.

### 4.3.

We now address a couple of delicate aspects of the criteria, giving a precise
description of the representation $\sigma^{\prime}$ and character $\chi$
appearing in § 4.1.

######  Proposition 4.3.

Let us assume that $\operatorname{Sym}^{4}(\sigma)$ is irreducible and that
$\operatorname{Sym}^{6}(\sigma)$ is reducible. Let $\sigma^{\prime}$ be the
representation in Theorem 4.1. Then there exists a unique character $\mu$ such
that

$\operatorname{Sym}^{3}(\sigma^{\prime})\cong\operatorname{Sym}^{3}(\sigma)\otimes\mu.$

More precisely, $\mu=\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1}\chi$, where
$\chi$ is as in Proposition 4.2, and satisfies

$\mu=(\eta\omega_{\sigma}^{2})^{3},$

for a quadratic character $\eta$.

###### Proof.

The unicity of $\mu$ follows from the irreducibility of
$\operatorname{Sym}^{4}(\sigma)$ and Theorem 3.5. Let $\chi$ be a character as
in Proposition 4.2. We recall that we have

$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\chi\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}.$

Then

$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\sigma\otimes{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\chi\cong\left(\operatorname{Sym}^{2}(\sigma)\otimes{\rm{Ad}}(\sigma^{\prime})\otimes\chi\right)\oplus\left({\rm{Ad}}(\sigma^{\prime})\otimes\chi\omega_{\sigma}\right),$

and

$\sigma\otimes\operatorname{Sym}^{5}(\sigma)\cong\sigma\otimes{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}\cong\left(A^{3}(\sigma)\otimes\sigma^{\prime}\right)\oplus\left(\sigma\otimes\sigma^{\prime}\right).$

Since $\sigma\otimes\sigma^{\prime}$ is irreducible, the identities above
imply that it is a subrepresentation of
$\operatorname{Sym}^{2}(\sigma)\otimes{\rm{Ad}}(\sigma^{\prime})\otimes\chi$.
Then $\chi$ is a character contained in
$\operatorname{Sym}^{2}(\sigma^{\vee})\otimes{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\sigma^{\prime}$.
Using Clebsch-Gordan, we see that $\chi$ is contained in

$\left(A^{3}(\sigma)\otimes
A^{3}(\sigma^{\prime})\otimes\omega_{\sigma}^{-1}\right)\oplus\left(A^{3}(\sigma)\otimes\sigma^{\prime}\otimes\omega_{\sigma}^{-1}\right)\oplus\left(\sigma^{\vee}\otimes
A^{3}(\sigma^{\prime})\right)\oplus\left(\sigma^{\vee}\otimes\sigma^{\prime}\right).$

The representation $\sigma\otimes\sigma^{\prime}$ being irreducible, the
character $\chi$ is not contained in $\sigma^{\vee}\otimes\sigma^{\prime}$.
Also, since $A^{3}(\sigma)$ and $\sigma^{\prime}$ are irreducible and of
different dimension, $\chi$ cannot be contained in
$A^{3}(\sigma)\otimes\sigma^{\prime}\otimes\omega_{\sigma}^{-1}$. Let us
assume that $\sigma^{\vee}\otimes A^{3}(\sigma^{\prime})$ contains $\chi$.
Then $A^{3}(\sigma^{\prime})$ is reducible and $\sigma\otimes\chi$ is one of
its irreducible factors. Now, since ${\rm{Ad}}(\sigma^{\prime})$ is
irreducible (otherwise,
$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma^{\prime})\otimes\sigma\otimes\chi$
would be reducible), then by Theorem 3.3 we have that

$A^{3}(\sigma^{\prime})\cong\sigma^{\prime}\otimes\mu_{0}\oplus\sigma^{\prime}\otimes\mu_{0}^{2},$

for some character $\mu_{0}$. Then $\sigma\otimes\chi$ is (isomorphic to)
either $\sigma^{\prime}\otimes\mu_{0}$ or $\sigma^{\prime}\otimes\mu_{0}^{2}$.
This is a contradiction, for we know that
$\sigma^{\prime}\not\cong\sigma\otimes\mu$ for every character $\mu$.

Therefore, $A^{3}(\sigma)\otimes
A^{3}(\sigma^{\prime})\otimes\omega_{\sigma}^{-1}$ contains $\chi$. This
implies that $A^{3}(\sigma^{\prime})$ is irreducible and
$A^{3}(\sigma^{\prime})\cong A^{3}(\sigma)^{\vee}\otimes\omega_{\sigma}\chi$.
Then if we let $\mu=\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1}\chi$, we get

$\operatorname{Sym}^{3}(\sigma^{\prime})\cong\operatorname{Sym}^{3}(\sigma)\otimes\mu,$

as desired.

Now, by comparing determinants from the relation
${\rm{Sym}}^{5}(\sigma)\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}$, we have

(4.5) $\omega_{\sigma}^{15}=\omega_{\sigma^{\prime}}^{3}.$

On the other hand, by taking contragradient from both sides of
${\rm{Sym}}^{3}(\sigma^{\prime})\cong{\rm{Sym}}^{3}(\sigma)\otimes\mu$, we
obtain
${\rm{Sym}}^{3}(\sigma^{\prime})\otimes\omega_{\sigma^{\prime}}^{-3}\cong{\rm{Sym}}^{3}(\sigma)\otimes\omega_{\sigma}^{-3}\mu^{-1}$.
Thus,

${\rm{Sym}}^{3}(\sigma)\otimes\mu\cong{\rm{Sym}}^{3}(\sigma^{\prime})\cong{\rm{Sym}}^{3}(\sigma)\otimes(\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1})^{3}\mu^{-1},$

i.e.

${\rm{Sym}}^{3}(\sigma)\cong{\rm{Sym}}^{3}(\sigma)\otimes(\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1})^{3}\mu^{-2}.$

Since $\sigma$ is not of octahedral type, the above implies that
$\mu^{2}=(\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1})^{3}$. But from (4.5)
we have
$(\omega_{\sigma^{\prime}}\omega_{\sigma}^{-1})^{3}=\omega_{\sigma}^{12}$.
Thus

$\mu=\eta\omega_{\sigma}^{6}=(\eta\omega_{\sigma}^{2})^{3}$

for some quadratic character $\eta$. ∎

We now register a consequence that we remarked in § 4.2.

######  Corollary 4.4.

Let $\sigma^{\prime}$ be the irreducible two-dimensional Galois representation
of Theorem 4.1 satisfying

$\operatorname{Sym}^{5}(\sigma)\cong{\rm{Ad}}(\sigma)\otimes\sigma^{\prime}.$

Then the character $\chi$ of Proposition 4.2 satisfying

$\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma^{\prime})\otimes\sigma\otimes\chi$

is unique with respect to this property.

###### Proof.

If $\xi$ were another character satisfying
$\operatorname{Sym}^{5}(\sigma)\cong{\rm
Ad}(\sigma^{\prime})\otimes\sigma\otimes\xi$, then Proposition 4.2 tells us
that

$\operatorname{Sym}^{6}(\sigma)\cong\sigma\otimes\sigma^{\prime}\oplus{\rm{Ad}}(\sigma^{\prime})\otimes\xi\omega_{\sigma}.$

Hence, we would have

${\rm{Ad}}(\sigma^{\prime})\cong{\rm Ad}(\sigma^{\prime})\otimes\xi\chi^{-1}.$

If $\xi\chi^{-1}$ were not trivial, then by Theorem 3.3,
$\operatorname{Sym}^{3}(\sigma^{\prime})$ would be reducible, a contradiction.
∎

### 4.4.

Let us denote $\xi:=\eta^{-1}\omega_{\sigma}^{-2}$ and
$\widetilde{\sigma}:=\sigma^{\prime}\otimes\xi$. Thus,
$\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\widetilde{\sigma})$.
We observe that this implies in particular that
$\operatorname{Sym}^{3}(\widetilde{\sigma})$ is irreducible. Also, by Theorem
3.5, we have that $\operatorname{Sym}^{4}(\widetilde{\sigma})$ is irreducible.
Furthermore, by decomposing via the Clebsch-Gordan formulas both sides of

$\operatorname{Sym}^{3}(\sigma)\otimes\operatorname{Sym}^{3}(\sigma)\cong\operatorname{Sym}^{3}(\widetilde{\sigma})\otimes\operatorname{Sym}^{3}(\widetilde{\sigma})$

and using the fact that $\operatorname{Sym}^{6}(\sigma)$ is reducible, we
deduce that $\operatorname{Sym}^{6}(\widetilde{\sigma})$ is reducible. As a
conclusion of this, $\widetilde{\sigma}$ satisfies the conditions in Theorem
4.1.

We write $\Gamma=\sigma(G_{F})$ and
$\widetilde{\Gamma}=\widetilde{\sigma}(G_{F})$. We shall need the following
useful fact.

######  Lemma 4.5.

The groups $\Gamma$ and $\widetilde{\Gamma}$ consist of semisimple
automorphisms.

###### Proof.

Let $g\in\Gamma$, and let us assume that $g$ is not semisimple. Then, the
Jordan decomposition tells us that there is a suitable basis for $V$ where we
may write

$g=\left(\begin{array}[]{cc}\lambda&1\\\ 0&\lambda\end{array}\right)$

for some $\lambda\in\overline{\mathbb{Q}}_{\ell}^{\times}$. But then the
Jordan normal form of $\operatorname{Sym}^{6}(g)$ has a single block, i.e.,
none of its invariant subspaces have a complement. This is not possible, since
$\operatorname{Sym}^{6}(\sigma)$ is semisimple and reducible. Since
$\operatorname{Sym}^{6}(\widetilde{\sigma})$ is also reducible, the same
applies to $\widetilde{\Gamma}$. ∎

In our study of the image $\Gamma=\sigma(G_{F})$, we come across the following
_no-small finite subgroups_ property:

> A compact subgroup $K$ of ${\rm GL}_{2}(\overline{\mathbb{Q}}_{\ell})$, has
> itself a compact open subgroup $\mathcal{U}$ without non-trivial finite
> subgroups.

Actually, if $K$ is any compact subgroup of ${\rm
GL}_{2}(\overline{\mathbb{Q}}_{\ell})$, then there exists a finite extension
of $\mathbb{Q}_{\ell}$ with ring of integers $\mathcal{O}$ such that
$K\subseteq{\rm GL}_{2}(\mathcal{O})$. It is well known that ${\rm
GL}_{2}(\mathcal{O})$ contains a compact open subgroup, say $\mathcal{U}_{0}$,
without non-trivial finite subgroups. Then it suffices to take
$\mathcal{U}=\mathcal{U}_{0}\cap K$.

We observe that $\mathcal{U}$ could very well be trivial, and clearly this is
the case when $K$ is finite. Note that $\Gamma=\sigma(G_{F})$ is compact,
hence it satisfies the no-small finite subgroups property.

######  Theorem 4.6.

If ${\rm{Sym}}^{4}(\sigma)$ is irreducible and ${\rm{Sym}}^{6}(\sigma)$ is
reducible, then $\sigma$ has open kernel and $J={\rm proj}(\sigma(G_{F}))$ is
icosahedral.

Our proof relies on the following technical lemma. That ${\rm
proj}(\sigma(G_{F}))$ is icosahedral follows once we settle that $\sigma$ has
open kernel, for we are assuming that $\sigma$ is neither dihedral,
tetrahedral, nor octahedral.

######  Lemma 4.7.

Assume that $\,\mathcal{U}_{\,\Gamma}$ is a non-trivial compact open subgroup
of $\,\Gamma=\sigma(G_{F})$ without non-trivial finite subgroups. Let

$\varphi\colon{\rm{Sym}}^{3}(\sigma)\xrightarrow{\ \sim\
}{\rm{Sym}}^{3}(\widetilde{\sigma})$

be a given $G_{F}$-isomorphism. Then there exist bases for $V$ and
$\widetilde{V}$ and $g\in{\rm{M}}_{2}(\overline{\mathbb{Q}}_{\ell})$ such that
the corresponding matrix for $\varphi$ is ${\rm{Sym}}^{3}(g)$.

###### Proof _(of Lemma 4.7)_.

Let us first assume that $\omega_{\sigma}$ is of finite order. Let
$h\in\mathcal{U}_{\,\Gamma}$ be different from the identity. We choose a basis
for $V$ such that $h$ is representable by a diagonal matrix, say
$h=\operatorname{diag}(a,b)$. Let us choose a basis for $\widetilde{V}$ such
that

$\varphi\cdot{\rm{Sym}}^{3}(h)\cdot\varphi^{-1}=\varphi\cdot\operatorname{diag}(a^{3},a^{2}b,ab^{2},b^{3})\cdot\varphi^{-1}\in{\rm{Aut}}_{\overline{\mathbb{Q}}_{\ell}}(\widetilde{V})$

is representable by a diagonal matrix. Since
$\varphi\colon{\rm{Sym}}^{3}(\sigma)\rightarrow{\rm{Sym}}^{3}(\widetilde{\sigma})$
is a $G_{F}$-isomorphism, we may write

(4.6)
$\varphi\cdot\operatorname{diag}(a^{3},a^{2}b,ab^{2},b^{3})\cdot\varphi^{-1}=\operatorname{diag}(\alpha^{3},\alpha^{2}\beta,\alpha\beta^{2},\beta^{3})$

for some $\alpha,\beta\in\overline{\mathbb{Q}}_{\ell}$. We note that all the
non-zero entries of ${\rm{Sym}}^{3}(h)$ are different from each other.
Otherwise we would have that $a^{k}=b^{k}$ for some integer $k>0$, and since
$h$ has a finite-order determinant, then $a^{n}=b^{n}=1$ for some other
integer $n>0$. But then $h$ would generate a finite non-trivial group,
contradicting the no-small finite subgroups property of $\Gamma$. Thus, we may
write $\varphi$ as $TM$, where $T$ is diagonal and $M$ a permutation matrix.
Let us see actually that $M$ is either $I=(\delta_{i,j})$ or
$I^{-}=(\delta_{i,5-j})$. From (4.6), we have

$\begin{split}\alpha^{3}&=a^{n_{0}}b^{3-n_{0}},\\\
\alpha^{2}\beta&=a^{n_{1}}b^{3-n_{1}},\\\
\alpha\beta^{2}&=a^{n_{2}}b^{3-n_{2}},\\\
\beta^{3}&=a^{n_{3}}b^{3-n_{3}},\end{split}$

for some permutation $(n_{0},n_{1},n_{2},n_{3})$ of $(0,1,2,3)$. Again, since
$a^{k}\not=b^{k}$ for all integers $k\not=0$, the equations above imply

$\begin{split}n_{0}+n_{3}&=n_{1}+n_{2},\\\ n_{0}+n_{2}&=2n_{1}.\end{split}$

It is readily seen that the unique solutions of the latter are $(0,1,2,3)$ and
$(3,2,1,0)$, as claimed.

Since both $I$ and $I^{-}$ are of the form ${\rm{Sym}}^{3}(g_{0})$ for some
$g_{0}$, it suffices to prove that $T={\rm{Sym}}^{3}(g)$ for some $g$. Let us
suppose that

$T=\operatorname{diag}(t_{1},t_{2},t_{3},t_{4}).$

Necessary and sufficient conditions for $T$ to be of the form
${\rm{Sym}}^{3}(g)$ for some $g$ are $t_{1}t_{4}=t_{2}t_{3}$ and
$t_{1}t_{3}=t_{2}^{2}$. Let us verify these. Since $\sigma$ is not dihedral,
there exists $\left(\begin{array}[]{cc}x&y\\\ w&z\\\ \end{array}\right)\in J$
such that $xy\not=0$ or $wz\not=0$. We note that
$T\cdot{\rm{Sym}}^{3}\left(\left(\begin{array}[]{cc}x&y\\\ w&z\\\
\end{array}\right)\right)\cdot T^{-1}$ is equal to

(4.7) $\left(\begin{array}[]{cccc}x^{3}&x^{2}y\cdot
t_{1}t_{2}^{-1}&xy^{2}\cdot t_{1}t_{3}^{-1}&y^{3}\cdot t_{1}t_{4}^{-1}\\\
&*&*&*\\\ &*&*&*\\\ w^{3}\cdot t_{4}t_{1}^{-1}&w^{2}z\cdot
t_{4}t_{2}^{-1}&wz^{2}\cdot t_{4}t_{3}^{-1}&z^{3}\end{array}\right),$

where we will only use the entries in the first and last row of the previous
matrix.

Let us suppose that $xy\not=0$. Since the matrix in (4.7) is
${\rm{Sym}}^{3}(\widetilde{h})$ for some $\widetilde{h}$, then we certainly
have

$\begin{split}x^{3}y^{3}\cdot t_{1}t_{4}^{-1}&=(x^{2}y\cdot
t_{1}t_{2}^{-1})(xy^{2}\cdot t_{1}t_{3}^{-1}),\\\ x^{3}(xy^{2}\cdot
t_{1}t_{3}^{-1})&=(x^{2}y\cdot t_{1}t_{2}^{-1})^{2}.\end{split}$

Since $xy\not=0$, then $t_{1}t_{4}^{-1}=t_{1}^{2}t_{2}^{-1}t_{3}^{-1}$ and
$t_{1}t_{3}^{-1}=t_{1}^{2}t_{2}^{-2}$. It is straightforward to see that these
lead us to the desired properties $t_{1}t_{4}=t_{2}t_{3}$ and
$t_{1}t_{3}=t_{2}^{2}$. If $wz\not=0$, then we must pay attention to the
fourth row of (4.7), and the above works similar.

In general, we note that, by [3] § IV.2.9, there exists an $\ell$-adic
character $\chi$ such that $\det(\sigma\otimes\chi)$ is of finite order. Since
every $G_{F}$-isomorphism

$\varphi\colon\rm{Sym}^{3}(\sigma\otimes\chi)\xrightarrow{\ \sim\
}{\rm{Sym}}^{3}(\widetilde{\sigma}\otimes\chi)$

is in turn a $G_{F}$-isomorphism

$\varphi\colon{\rm{Sym}}^{3}(\sigma)\xrightarrow{\ \sim\
}{\rm{Sym}}^{3}(\widetilde{\sigma}),$

the proposition follows. ∎

###### Proof of Theorem 4.6.

Let $\mathcal{U}_{\,\Gamma}$ be a compact open subgroup of $\Gamma$ without
non-trivial finite subgroups. If $\mathcal{U}_{\,\Gamma}$ is trivial, then
$\Gamma$ is finite and we are done. Hence, we may assume that
$\mathcal{U}_{\,\Gamma}$ is non-trivial.

We fix bases for $V$ and $\widetilde{V}$ and choose
$g\in{\rm{Hom}}_{\overline{\mathbb{Q}}_{\ell}}(V,\widetilde{V})$ such that
${\rm{Sym}}^{3}(g)$ is a $G_{F}$-isomorphism
${\rm{Sym}}^{3}(\sigma)\xrightarrow{\ \sim\
}{\rm{Sym}}^{3}(\widetilde{\sigma})$, just as in Lemma 4.7. Since the group
homomorphism
${\rm{Sym}}^{3}\colon{\rm{GL}}_{2}(\overline{\mathbb{Q}}_{\ell})\to{\rm{GL}}_{4}(\overline{\mathbb{Q}}_{\ell})$
has finite kernel, then there exists an open subgroup $U$ of $\Gamma$ such
that ${\rm{Sym}}^{3}|_{U}$ is injective.

If we set $H=\sigma^{-1}(U)$, then $H$ is an open subgroup of $G_{F}$ and $g$
becomes an $H$-isomorphism $V\xrightarrow{\ \sim\ }\widetilde{V}$. We observe
that since
$\sigma_{H}\cong\widetilde{\sigma}_{H}\cong\sigma^{\prime}_{H}\otimes\xi_{H}$,
and

${\rm{Sym}}^{5}(\sigma_{H})\cong{\rm{Ad}}(\sigma_{H})\otimes\sigma^{\prime}_{H},$

then ${\rm{Sym}}^{5}(\sigma_{H})$ is reducible. This implies that $\sigma_{H}$
has open kernel. Since $H$ is open in $G_{F}$, we are done. ∎

###  4.5. Higher Symmetric Powers

We here register Ramakrishnan’s irreducibility criterion for higher symmetric
powers, Theorem A’ (c) of [15].

######  Theorem 4.8.

Let $n>6$ be an integer. Then $\operatorname{Sym}^{n}(\sigma)$ is irreducible
if and only if $\operatorname{Sym}^{6}(\sigma)$ is irreducible.

The proof for complex representations over number fields of [_loc. cit._], is
also valid in the case of $\ell$-adic representations over function fields. It
is a formal consequence of Lemmas 2.1 and 2.2.

### 4.6.

The following theorem summarizes what we have done so far.

######  Theorem 4.9.

Let $\sigma\colon G_{F}\to{\rm GL}(V)$ be an irreducible 2-dimensional
$\ell$-adic Galois representation. Then the following are equivalent:

  * (i)

$\sigma$ has open kernel.

  * (ii)

There exists an integer $n\geq 2$ such that $\operatorname{Sym}^{n}(\sigma)$
is reducible.

  * (iii)

$\operatorname{Sym}^{6}(\sigma)$ is reducible.

To be more precise, if the above equivalent conditions are met we then have
the following classification.

  * M=1: $\sigma$ is dihedral.

  * M=2: $\sigma$ is tetrahedral.

  * M=3: $\sigma$ is octahedral.

  * M=5: $\sigma$ is icosahedral.

And, $\operatorname{Sym}^{5}(\sigma)$ is irreducible if and only if
$\operatorname{Sym}^{4}(\sigma)$ is irreducible.

###### Proof.

At this point, only the equivalence of (i), (ii) and (iii) require comment.

Assume that $\sigma$ has open kernel, then $\Gamma=\sigma(G_{F})$ is finite.
Now, for each $n$, $\operatorname{Sym}^{n}(\sigma)$ can be seen as a
representation of $\Gamma$ over the algebraically closed field of
characteristic zero $\overline{\mathbb{Q}}_{\ell}$. Notice that $\Gamma$ can
only have finitely many irreducible representations up to isomorphism.
Necessarily, $\operatorname{Sym}^{n}(\sigma)$ is reducible for some integer
$n\geq 2$ and (ii) holds.

Now, assume that $\operatorname{Sym}^{n}(\sigma)$ is reducible for some $n\geq
2$. If $n\leq 6$, then $\operatorname{Sym}^{6}(\sigma)$ is reducible by Lemma
2.1. And, if $n>6$ Theorem 4.8 implies that $\operatorname{Sym}^{6}(\sigma)$
is reducible.

Finally, assume that $\operatorname{Sym}^{6}(\sigma)$ is reducible. Then the
maximal irreducible symmetric power satisfies $M\leq 5$. The case of $M=4$
cannot happen by Lemma 2.2, and each case among $M=1,2,3$ is covered by one of
the irreducibility criteria of § 3 and Theorem 4.6 applies to $M=5$. It
follows that $\sigma$ has open kernel. ∎

##  5\. Passage to Automorphic Representations

The landmark results of L. Lafforgue over a global function field $F$ [11],
establish a correspondence between cuspidal automorphic representations of
${\rm GL}_{n}(\mathbb{A}_{F})$ with finite central character and irreducible
$n$-dimensional $\ell$-adic Galois representations that are unramified almost
everywhere and have finite determinant.

In [3], the global Langlands correspondence is slightly expanded. It is
phrased in the context of the more general $\ell$-adic Weil representations,
which we have taken to be unramified almost everywhere by definition in § 1.
The crucial point in this generalization is that for an arbitrary $\ell$-adic
Weil representation $\sigma\colon\mathcal{W}_{F}\to{\rm GL}(V)$, there exists
an $\ell$-adic character $\chi$ such that $\sigma\otimes\chi$ can be extended
to an $\ell$-adic Galois representation with finite-order determinant. That is
to say, $\sigma\otimes\chi$ lies in the setting of L. Lafforgue [11]. More
precisely, in [3] the authors establish a bijection between the set
$\mathcal{G}^{n}_{\ell}(F)$ of irreducible $n$-dimensional $\ell$-adic
representations of $\mathcal{W}_{F}$ and the set $\mathcal{A}^{n}(F)$ of
cuspidal automorphic representations of ${\rm GL}_{n}(\mathbb{A}_{F})$, such
that if $\sigma\in\mathcal{G}^{n}_{\ell}(F)$ corresponds to
$\pi\in\mathcal{A}^{n}(F)$, then it agrees at every place with the local
Langlands correspondence established by Laumon, Rapoport and Stuhler [13]. We
write $\sigma\longleftrightarrow\pi$ to denote corresponding representations.

An $\ell$-adic Galois character $\chi$ corresponds to an automorphic character
which we again denote in this case by $\chi$, this is possible via class field
theory after fixing a field isomorphism
$\iota\colon\overline{\mathbb{Q}}_{\ell}\to\mathbb{C}$, cf. § IV.2 of [3].
Furthermore, if $\pi\in\mathcal{A}^{n}(F)$ corresponds to
$\sigma\in\mathcal{G}^{n}_{\ell}(F)$, then the central character
$\omega_{\pi}$ of $\pi$ corresponds to $\omega_{\sigma}=\det(\sigma)$.

The correspondence is further extended so that a semisimple $n$-dimensional
$\ell$-adic representation $\sigma$ of $\mathcal{W}_{F}$, written as a direct
sum

$\sigma=\sigma_{1}\oplus\cdots\oplus\sigma_{d},\quad\sigma_{i}\in\mathcal{G}^{n_{i}}_{\ell}(F),$

corresponds to an automorphic representation $\pi$ of ${\rm
GL}_{n}(\mathbb{A}_{F})$, given by an isobaric sum

$\pi=\pi_{1}\boxplus\cdots\boxplus\pi_{d},\quad\pi_{i}\in\mathcal{A}^{n_{i}}(F).$

The correspondence is such that $\sigma_{i}\longleftrightarrow\pi_{i}$, for
each $i=1,\ldots,d$; and, we have equality of Artin and automorphic
$L$-functions

$L(s,\sigma_{i})=L(s,\pi_{i}).$

This latter form of the Langlands correspondence between semisimple
representations on the Galois side and automorphic representations of the
general linear group is the one we consider; we again write
$\sigma\longleftrightarrow\pi$ in this setting for corresponding
representations.

### 5.1.

We are fortunate in the case of function fields that we can extend Theorem 4.9
to irreducible $\ell$-adic representations of $\mathcal{W}_{F}$ and pass to
the automorphic side of the global Langlands correspondence to obtain Theorem
5.1 below. The general classification result on the automorphic side can be
succintly stated if we let $M$ be the maximal cuspidal symmetric power of
$\pi$. When $M$ exists as a finite positive integer, it is determined by ${\rm
Sym}^{M}(\pi)$ being cuspidal, while ${\rm Sym}^{M+1}(\pi)$ is non-cuspidal.
We write $M=\infty$ in case every symmetric power of $\pi$ is cuspidal.

######  Theorem 5.1.

Let $\pi$ be a cuspidal automorphic representation of ${\rm
GL}_{2}(\mathbb{A}_{F})$. Then ${\rm Sym}^{6}(\pi)$ is cuspidal if and only if
$M=\infty$. If ${\rm Sym}^{6}(\pi)$ is non-cuspidal, then $\pi$ admits the
following classification.

  * M=1: $\pi$ is dihedral.

  * M=2: $\pi$ is tetrahedral.

  * M=3: $\pi$ is octahedral.

  * M=5: $\pi$ is icosahedral.

Additionally, ${\rm Sym}^{4}(\pi)$ is cuspidal if and only ${\rm
Sym}^{5}(\pi)$ is cuspidal.

###### Proof.

Let $\sigma\colon\mathcal{W}_{F}\to{\rm GL}(V)$ be the irreducible
2-dimensional $\ell$-adic representation attached to $\pi$. Let $\chi$ be a
character such that $\sigma\otimes\chi$ can be extended to $G_{F}$.

We continue to denote the extension of $\sigma\otimes\chi$ to $G_{F}$ by
$\sigma\otimes\chi$. For every integer $n\geq 2$, we consider the
representations

$\operatorname{Sym}^{n}(\sigma\otimes\chi)=\operatorname{Sym}^{n}(\sigma)\otimes\chi^{n}\colon
G_{F}\to{\rm GL}(\operatorname{Sym}^{n}(V)),$

and

$\operatorname{Sym}^{n}(\sigma)\colon\mathcal{W}_{F}\to{\rm
GL}(\operatorname{Sym}^{n}(V)).$

Observe that $\operatorname{Sym}^{n}(\sigma\otimes\chi)$ is reducible if and
only if $\operatorname{Sym}^{n}(\sigma)$ is reducible.

If $\operatorname{Sym}^{n}(\sigma)$ is reducible for some $n$ then
$\sigma\otimes\chi$ has open kernel, by Theorem 4.9; in particular, ${\rm
proj}((\sigma\otimes\chi)(G_{F}))$ is finite. Since there is an inclusion map
$\mathcal{W}_{F}\hookrightarrow G_{F}$ with dense image and the involved
representations to ${\rm PGL}_{2}(\overline{\mathbb{Q}}_{\ell})$ are
continuous, ${\rm proj}((\sigma\otimes\chi)(\mathcal{W}_{F}))={\rm
proj}(\sigma(\mathcal{W}_{F}))$ is forced to equal ${\rm
proj}((\sigma\otimes\chi)(G_{F}))$.

With these observations, the theorem follows from Theorem 4.9, via the global
Langlands correspondence. ∎

Over number fields, the cuspidality criterion for symmetric powers is
conjectured in [9], see § 3 Conjecture therein; one can there find proofs of
the characteristic zero statements corresponding to the cases $M=2,3$ of
Theorem 5.1.

Let us next proceed to the automorphic statements corresponding to the
detailed criteria obtained along the course of proving our results on the
Galois side. In the refined statements, we make use of quadratic base change
and automorphic induction, available to us in greater generality thanks to the
work of Henniart-Lemaire [3].

###  5.2. On cuspidality criteria for the symmetric square, cube and fourth

We work with a cuspidal representation $\pi$ of ${\rm
GL}_{2}(\mathbb{A}_{F})$, and make use of the following notation:

$A^{n}(\pi)=\operatorname{Sym}^{n}(\pi)\otimes\omega_{\pi}^{-1},$

where $\omega_{\pi}$ denotes the central character of $\pi$. The case of $n=2$
is the adjoint lift of Gelbart-Jacquet [2], which we denote by ${\rm
Ad}(\pi)$.

To begin with, in the case of symmetric square, Galois Theorem 3.1 allows us
to infer that the following statements are equivalent:

  1. (i)

$\operatorname{Sym}^{2}(\pi)$ is non-cuspidal.

  2. (ii)

$\pi\cong\pi\otimes\chi$ for some non-trivial character $\chi$.

Note that $\pi$ is dihedral precisely when these conditions are met, by
Theorem 5.1.

From Galois Theorem 3.3, translated to the automorphic side: if $\pi$ is non-
dihedral, then $\operatorname{Sym}^{3}(\pi)$ is non-cuspidal if and only if
there exists some non-trivial character $\mu$ such that

$\operatorname{Sym}^{2}(\pi)\cong\operatorname{Sym}^{2}(\pi)\otimes\mu;$

the condition being equivalent to $\pi$ being tetrahedral. When this is the
case we have the decomposition

$A^{3}(\pi)\cong\pi\otimes\mu\boxplus\pi\otimes\mu^{2}.$

Now, assume that $\operatorname{Sym}^{3}(\pi)$ is irreducible. Via Galois
Theorem 3.5, we obtain that $\operatorname{Sym}^{4}(\pi)$ is cuspidal if and
only if there exists a non-trivial quadratic character $\chi$ such that

$\operatorname{Sym}^{3}(\pi)\cong\operatorname{Sym}^{3}(\pi)\otimes\chi;$

the condition being equivalent to $\pi$ being octahedral. When this is the
case, we get

$A^{4}(\pi)\cong\Pi_{E}^{F}(\omega_{\pi_{E}}\mu)\boxplus\operatorname{Sym}^{2}(\pi)\otimes\chi,$

where $E/F$ is the quadratic extension corresponding to $\chi$ via class field
theory and $\mu$ is some cubic character of $\mathbb{A}_{E}^{\times}$. Let us
explain the notation, where we are assuming $\pi\longleftrightarrow\sigma$,
i.e., $\pi\in\mathcal{A}^{2}(F)$ corresponds to
$\sigma\in\mathcal{G}^{2}_{\ell}(F)$ under global Langlands; with
corresponding base change $\pi_{E}\longleftrightarrow\sigma_{E}$. Then we have
induced representations on the Galois side and monomial representations on the
automorphic side, in particular,

$\rm{Ind}_{E}^{F}(\omega_{\sigma_{E}}\mu)\longleftrightarrow\Pi_{E}^{F}(\omega_{\pi_{E}}\mu),$

where $\mu$ is some cubic character of $G_{E}$ and $\Pi_{E}^{F}$ denotes
automorphic induction, cf. § IV.5 of [3].

###  5.3. On cuspidality criteria involving symmetric sixth

If $\operatorname{Sym}^{4}(\pi)$ is cuspidal, then Galois Theorem 4.1 leads us
to conclude that the following statements are equivalent:

  1. (i)

$\operatorname{Sym}^{6}(\pi)$ is non-cuspidal.

  2. (ii)

There exists a cuspidal two-dimensional representation $\pi^{\prime}$ of ${\rm
GL}_{2}(\mathbb{A}_{F})$ such that
$\operatorname{Sym}^{5}(\pi)\cong{\rm{Ad}}(\pi)\boxtimes\pi^{\prime}$.

The representation $\pi^{\prime}$ is uniquely determined up to isomorphism by
(ii).

Next, we look into the results corresponding to Galois Proposition 4.2. Assume
that $\operatorname{Sym}^{4}(\pi)$ is cuspidal and
$\operatorname{Sym}^{6}(\pi)$ is non-cuspidal. Then there exists a unique
character $\chi$ such that

$\operatorname{Sym}^{5}(\pi)\cong{\rm
Ad}(\pi^{\prime})\boxtimes\pi\otimes\chi,$

and we have the decomposition

$\operatorname{Sym}^{6}(\pi)\cong\pi\boxtimes\pi^{\prime}\boxplus{\rm{Ad}}(\pi^{\prime})\otimes\chi\omega_{\pi}.$

Notice that the uniqueness of $\chi$ is obtained via Galois Corollary 4.4.
Furthermore, from Galois Proposition 4.3, we obtain that

$\operatorname{Sym}^{3}(\pi^{\prime})\cong\operatorname{Sym}^{3}(\pi)\otimes\mu,$

for $\mu=\omega_{\pi^{\prime}}\omega_{\pi}^{-1}\chi$; and, one can also write
$\mu=(\eta\,\omega_{\pi}^{2})^{3}$ for a quadratic character $\eta$.

##  Appendix A L-functions and Subrepresentations

In this paragraph, we gather results about Artin $L$-functions and
reducibility over a global function field $F$, and formulate a useful property
on subrepresentations. In view of the global Langlands correspondence of L.
Lafforgue [11] summarized in § 5 following Henniart-Lemaire [3], and noting
that an $\ell$-adic representation of $G_{F}$ restricted to $\mathcal{W}_{F}$
is an $\ell$-adic representation, it is enough for us to work with semisimple
$\ell$-adic representations of $G_{F}$. Finite dimensional Galois
representations correspond to automorphic representations of general linear
groups. Now, on the automorphic side of the correspondence we work with
Rankin-Selberg products and their $L$-functions, where further references can
be found in [4].

Let $\sigma\colon G_{F}\to\text{GL}(V)$ be a given semisimple $\ell$-adic
$n$-dimensional representation. We can write $\sigma$ as the direct sum

$\sigma=\sigma_{1}\oplus\cdots\oplus\sigma_{d},$

where each $\sigma_{i}$ is an irreducible representation of dimension $n_{i}$,
for $i=1,\ldots,d$. We note that each $\sigma_{i}$ is an $\ell$-adic
representation, since they are all continuous and unramified at each place
where $\sigma$ is unramified. For each $i=1,\ldots,d$ there exists a cuspidal
automorphic representation $\pi_{i}$ of ${\rm GL}_{n_{i}}(\mathbb{A}_{F})$
such that the automorphic representation of ${\rm GL}_{n}(\mathbb{A}_{F})$
given by

$\pi=\pi_{1}\boxplus\cdots\boxplus\pi_{d},$

is such that $\pi\longleftrightarrow\sigma$, i.e., $\pi$ corresponds to
$\sigma$ under global Langlands.

Given semisimple $\ell$-adic Galois representations $\sigma$ and
$\sigma^{\prime}$ of finite dimensions $n$ and $n^{\prime}$, respectively,
Artin $L$-functions satisfy the following additive property

$L(s,\sigma\oplus\sigma^{\prime})=L(s,\sigma)L(s,\sigma^{\prime}).$

For automorphic representations $\pi$ of ${\rm GL}_{n}(\mathbb{A}_{F})$ and
$\pi^{\prime}$ of ${\rm GL}_{n^{\prime}}(\mathbb{A}_{F})$, written as isobaric
sums

$\pi=\pi_{1}\boxplus\cdots\boxplus\pi_{d},\quad\pi^{\prime}=\pi_{1}^{\prime}\boxplus\cdots\boxplus\pi_{e}^{\prime},$

we have the following multiplicativity property of Rankin-Selberg
$L$-functions

$L(s,\pi\times\pi^{\prime})=\prod_{i=1}^{d}\prod_{j=1}^{e}L(s,\pi_{i}\times\pi_{j}^{\prime}).$

If the Galois and automorphic representations are such that they correspond to
each other under global Langlands

$\sigma\longleftrightarrow\pi,\quad\sigma^{\prime}\longleftrightarrow\pi^{\prime},$

then Artin $L$-functions and Rankin-Selberg products are related by

$L(s,\sigma\otimes\sigma^{\prime})=L(s,\pi\times\pi^{\prime}).$

Now, fix an automorphic representation $\pi$ and a semisimple Galois
representation $\sigma$ that correspond to each other under global Langlands

$\pi=\pi_{1}\boxplus\cdots\boxplus\pi_{d}\ \longleftrightarrow\
\sigma=\sigma_{1}\oplus\cdots\oplus\sigma_{d}.$

Considering the contragredient representation

$\pi^{\vee}=\pi_{1}^{\vee}\boxplus\cdots\boxplus\pi_{d}^{\vee},$

we find that

$L(s,\pi\times\pi^{\vee})=\prod_{i=1}^{d}\prod_{j=1}^{d}L(s,\pi_{i}\times\pi_{j}^{\vee}).$

Since
$L(s,\pi_{i}\times\pi_{j}^{\vee})=L(s,\sigma_{i}\otimes\sigma_{j}^{\vee})$,
$1\leq i,j\leq d$, the additive property of Artin $L$-functions leads us to

$L(s,\pi\times\pi^{\vee})=\prod_{i=1}^{d}\prod_{j=1}^{d}L(s,\sigma_{i}\otimes\sigma_{j}^{\vee})=L(s,\sigma\otimes\sigma^{\vee}).$

Next, we employ a couple of facts about automorphic $L$-functions. Namely, a
non-vanishing result of Shahidi [17] and properties for Rankin-Selberg
products found in §§ 3 and 4 of [6] part II:
$L(s,\pi_{i}\times\pi_{j}^{\vee})$ does not vanish at $s=1$; and, it has a
simple pole at $s=1$ if $\pi_{j}\cong\pi_{i}$, otherwise it is invertible at
$s=1$. In this manner, we derive that the representation $\sigma$ is
irreducible if and only if $L(s,\sigma\otimes\sigma^{\vee})$ has a simple pole
at $s=1$. Additionally, if $\sigma$ is irreducible and $\tau$ is another
irreducible finite-dimensional $\ell$-adic representation, we then have
$L(s,\sigma\otimes\tau)$ is invertible at $s=1$, unless
$\tau\cong\sigma^{\vee}$, in which case $L(s,\sigma\otimes\tau)$ has a simple
pole at $s=1$. We summarize with the following general property:

  1. (A.1)

Given $\sigma$ semisimple and $\tau$ irreducible finite-dimensional
$\ell$-adic representations of $G_{F}$, then $L(s,\sigma\otimes\tau^{\vee})$
has a pole at $s=1$ if and only if $\tau$ appears as a subrepresentation of
$\sigma$. The multiplicity of $\tau$ in $\sigma$ is $-{\rm
ord}_{s=1}L(s,\sigma\otimes\tau^{\vee})$.

In fact, one immediately sees that if $\sigma$ and $\tau$ are in general
semisimple $\ell$-adic Galois representations of finite dimension such that
$\tau$ is a subrepresentation of $\sigma$, then
$L(s,\sigma\otimes\tau^{\vee})$ has a pole at $s=1$.

Now, suppose that $\rho,\sigma$ and $\tau$ are semisimple finite-dimensional
$\ell$-adic Galois representations, with $\sigma$ irreducible. If $\tau$ is a
subrepresentation of $\rho\otimes\sigma$, then
$L(s,\rho\otimes\sigma\otimes\tau^{\vee})$ has a pole at $s=1$, by Property
(A.1). But

$L(s,\rho\otimes\sigma\otimes\tau^{\vee})=L(s,\rho\otimes\tau^{\vee}\otimes(\sigma^{\vee})^{\vee}),$

then $\sigma^{\vee}$ is a subrepresentation of $\rho\otimes\tau^{\vee}$, again
by Property (A.1). For reference purposes, we register the following
represention-theoretical property:

  1. (A.2)

Let $\rho,\sigma$ and $\tau$ be semisimple finite-dimensional $\ell$-adic
representations of $G_{F}$, with $\sigma$ irreducible. If $\tau$ is a
subrepresentation of $\rho\otimes\sigma$, then $\sigma^{\vee}$ is a
subrepresentation of $\rho\otimes\tau^{\vee}$; equivalently, $\sigma$ is a
subrepresentation of $\rho^{\vee}\otimes\tau$.

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Luis Alberto Lomelí, Instituto de Matemáticas, Pontificia Universidad Católica
de Valparaíso, Blanco Viel 596, Cerro Barón, Valparaíso, Chile
_E-mail address:_<EMAIL_ADDRESS>
Francisco Javier Navarro, Instituto de Matemáticas, Pontificia Universidad
Católica de Valparaíso, Blanco Viel 596, Cerro Barón, Valparaíso, Chile
_E-mail address:_<EMAIL_ADDRESS>