# Wild Galois representations: Elliptic curves over a $3$-adic field

Nirvana Coppola  University of Bristol
School of Mathematics, Fry Building
Woodland Road
BS8 1UG Bristol, UK<EMAIL_ADDRESS>
###### Abstract.

Given an elliptic curve $E$ over a local field $K$ with residue characteristic
$3$, we investigate the action of the absolute Galois group of $K$ in the case
of potentially good reduction. The hardest case is when the $\ell$-adic Galois
representation attached to $E$ has non-cyclic inertia image, isomorphic to
$C_{3}\rtimes C_{4}$. In this work we describe such a representation
explicitly.

###### Key words and phrases:

Elliptic curves over local fields, Galois representations, non-Abelian inertia
action

###### 2010 Mathematics Subject Classification:

Primary 11G07; Secondary 11F80

##  1\. Introduction

Let $E$ be an elliptic curve defined over a field $K$, i.e. a projective
smooth curve of genus $1$, with a fixed point $O$. It is given by a non-
singular Weierstrass equation

$\displaystyle y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6},$

with coefficients $a_{i}\in K$. We refer to [S86, III §1] for the definition
of the invariants associated to this curve, in particular we denote by
$\Delta$ its discriminant. We denote by $\overline{K}$ a fixed algebraic
closure of $K$, and let $G_{K}=\operatorname{Gal}(\overline{K}/K)$ be the
absolute Galois group of $K$; then any $\sigma\in G_{K}$ acts on the points of
the curve by sending the point $P$ of coordinates $(x,y)$ to
$\sigma(P)=(\sigma(x),\sigma(y))$. In particular, for any integer $m$, $G_{K}$
acts on the group of $m$-torsion points $E[m]$ (i.e. the points of order
dividing $m$). This induces, for any prime $\ell$, an action on the
$\ell$-adic Tate modules, which are defined as

$\displaystyle T_{\ell}(E)=\varprojlim_{n}E[\ell^{n}].$

This action is called $\ell$-adic Galois representation attached to $E$, and
denoted $\rho_{E,\ell}$. For $\ell$ different from the characteristic of $p$,
we have that $T_{\ell}(E)$ is a free $\mathbb{Z}_{\ell}$-module of rank $2$;
so after considering the tensor product with an algebraic closure
$\overline{\mathbb{Q}}_{\ell}$ of $\mathbb{Q}_{\ell}$, we view
$\operatorname{Aut}(T_{\ell}(E))$ as a subgroup of
$\operatorname{GL}_{2}(\overline{\mathbb{Q}}_{\ell})$, and we also denote by
$\rho_{E,\ell}$ the representation we obtain:

$\displaystyle\rho_{E,\ell}:G_{K}\rightarrow\operatorname{GL}_{2}(\overline{\mathbb{Q}}_{\ell}).$

Furthermore, we fix an embedding
$\overline{\mathbb{Q}}_{\ell}\rightarrow\mathbb{C}$, and in what follows we
identify the eigenvalues and traces of the elements of $G_{K}$ with the
corresponding complex numbers.

Let $K$ be a local field of characteristic $0$, complete with respect to a
discrete valuation $v$, with valuation ring $O_{K}$, maximal ideal
$\mathfrak{m}$ and perfect residue field $k$. The residue characteristic of
$K$ is the characteristic of $k$, and we will always denote it by $p$. A
typical example of this is $K=\mathbb{Q}_{p}$, with the $p$-adic valuation, or
any algebraic extension of it. We denote by $K^{nr}$ the maximal unramified
extension of $K$ contained in $\overline{K}$.

Recall that the inertia subgroup $I_{K}$ of $G_{K}$ is isomorphic to
$\operatorname{Gal}(\overline{K}/K^{nr})$ and, if we denote by $\overline{k}$
a fixed algebraic closure of $k$, we have

$\displaystyle G_{K}/I_{K}\cong\operatorname{Gal}(\overline{k}/k).$

This group is procyclic, generated by the Frobenius element, i.e. the
automorphism: $x\mapsto x^{|k|}$ for all $x\in\overline{k}$. Any preimage
$\operatorname{Frob}$ of the Frobenius in $G_{K}$ under the projection to
$\operatorname{Gal}(\overline{k}/k)$ is called a Frobenius element of $G_{K}$.

To study the $\ell$-adic Galois representation attached to an elliptic curve
one has to study the Frobenius and the inertia actions and see how they are
related. This representation changes substantially according to the reduction
type of the curve, which can be good, multiplicative, or additive (for the
definition see [S86, VII §5]); moreover in the last case, there is a finite
extension of $K$ where the curve acquires either good or multiplicative
reduction, so we say $E/K$ has potentially good (resp. multiplicative)
reduction, as explained in Theorem 2.1. Let $p$ and $\ell$ be two distinct
primes, and suppose that the field $K$ over which $E$ is defined is a $p$-adic
field, i.e. isomorphic to a finite extension of $\mathbb{Q}_{p}$. Then the
$\ell$-adic Galois representation attached to $E$ can easily be studied in the
case of good and potentially multiplicative reduction, respectively via the
Criterion of Néron-Ogg-Shafarevich (see [S86, VII §7] and Theorem 2.3) and the
theory of the Tate curve (see [S94, V §3]); the last remaining case is that of
bad additive reduction with potentially good reduction. By the work of Kraus
([K90]) we have that the image of inertia is isomorphic to one of the
following:

$\displaystyle C_{2},C_{3},C_{4},C_{6},$ $\displaystyle C_{3}\rtimes$
$\displaystyle C_{4}\text{ only when }p=3,$ $\displaystyle Q_{8},SL_{2}$
$\displaystyle(\mathbb{F}_{3})\text{ only when }p=2,$

where $C_{n}$ denotes the cyclic group of order $n$ and $Q_{8}$ is the
quaternion group.

In this work we present a result to determine the $\ell$-adic Galois
representation $\rho_{E,\ell}$ attached to an elliptic curve over a field of
residue characteristic $3$, where $\ell$ is a prime different from $3$, such
that the image of inertia is non-cyclic, hence isomorphic to $C_{3}\rtimes
C_{4}$. We will use the notation of [GN] for group names and presentations and
for their character tables. In particular we will prove the following result.

######  Theorem 1.1.

Let $E$ be an elliptic curve with potentially good reduction over a $3$-adic
field $K$, with Weierstrass equation of the form $y^{2}=f(x)$ and discriminant
$\Delta$. Fix a fourth root $\Delta^{1/4}$ of $\Delta$ and define $F$ to be
the compositum of the splitting field of $f$ over $K$ and $K(\Delta^{1/4})$;
let $F^{\prime}$ be the Galois closure of $F/K$.

Let $\chi$ be the unramified character of $G_{K}$ (i.e. trivial on inertia)
such that

$\displaystyle\chi(\operatorname{Frob})=i^{n}\sqrt{3^{n}},$

and let $\psi$ be as follows. Let $n=[k:\mathbb{F}_{3}]$. If $n$ is even, let
$\psi$ be the representation of $\operatorname{Gal}(F^{\prime}/K)$, which is
isomorphic to $C_{3}\rtimes C_{4}$, with character:

$\displaystyle\begin{array}[]{c|rrrrrr}\rm class&\rm 1&\rm 2&\rm 3&\rm 4A&\rm
4B&\rm 6\cr\rm size&1&1&2&3&3&2\cr\hline\cr\psi&2&-2&-1&0&0&1\cr\end{array}$

while if $n$ is odd then let $\psi$ be the representation of
$\operatorname{Gal}(F^{\prime}/K)$, which is isomorphic to $C_{3}\rtimes
D_{4}$, with character:

$\displaystyle\begin{array}[]{c|rrrrrrrrr}\rm class&\rm 1&\rm 2A&\rm 2B&\rm
2C&\rm 3&\rm 4&\rm 6A&\rm 6B&\rm 6C\cr\rm
size&1&1&2&6&2&6&2&2&2\cr\hline\cr\psi&2&-2&0&0&-1&0&-i\sqrt{3}&i\sqrt{3}&1\cr\end{array}$

where the presentation of $\operatorname{Gal}(F^{\prime}/K)$ and its conjugacy
classes are as in Section 3.1.

Let $\ell$ be a prime different from $3$, and let $\rho_{E,\ell}$ be the
$\ell$-adic Galois representation attached to $E$. Suppose that the image of
the inertia subgroup of $G_{K}$ is isomorphic to $C_{3}\rtimes C_{4}$. Then
$\rho=\rho_{E,\ell}$ factors as

$\displaystyle\rho=\chi\otimes\psi.$

Part of this result follows immediately from [K90, Theorem 1], [ST68, Theorem
2] and the classification of the representations of the groups $C_{3}\rtimes
C_{4}$ and $C_{3}\rtimes D_{4}$ in [GN]. The result for the case of odd $n$ is
more subtle, as there are two possibilities for $\psi$, and we will prove that
only one of these occurs, via explicit computation.

In Section 2 we present the general setting for this problem. In particular we
recall some results about the Galois representations attached to elliptic
curves with good reduction and what is already known about the potentially
good reduction case. In Section 3 we tackle the case over $3$-adic fields when
the action of inertia is non-cyclic. In Section 3.1, we fix the a presentation
for the group $\operatorname{Gal}(F^{\prime}/K)$. The proof is divided into
two parts: in Section 3.2 we give the proof for the case of even $n$, and in
Section 3.3 we assume that $n$ is odd.

##  2\. Preliminaries

Let $E$ be an elliptic curve over a $p$-adic field $K$. In this section we
give some general results about the study of the $\ell$-adic Galois
representation attached to $E$, for $\ell\neq p$. As mentioned in the
introduction, this representation varies according to the different reduction
types that $E$ can have. There are several results about reduction types in
[S86, VII §5], which can be summarised as follows.

######  Theorem 2.1.

  * •

Let $K^{\prime}/K$ be an unramified extension. Then the reduction types of
$E/K$ and $E/K^{\prime}$ are the same.

  * •

Let $K^{\prime}/K$ be a finite extension. If $E/K$ has good or multiplicative
reduction over $K$, then $E/K^{\prime}$ has the same reduction type.

  * •

There exists a finite extension $K^{\prime}/K$ over which $E$ has either good
or multiplicative reduction.

  * •

The curve $E$ has potentially good reduction if and only if its $j$-invariant
is integral.

So if $E/K$ has additive reduction, then there is a finite extension over
which it acquires either good or multiplicative reduction; once it does, its
reduction type does not change after a finite extension. This suggests the
terminology _potentially good (resp. multiplicative) reduction_. In case of
additive reduction, using the last statement, we can check easily whether the
reduction is potentially good or not: we only need to compute the valuation of
the $j$-invariant.

######  Definition 2.2.

A Galois representation is _unramified_ if it is trivial when restricted to
the inertia subgroup.

We recall here the main content of the Criterion of Néron - Ogg - Shafarevich.
For the full statement and the proof see [S86, VII §7 Theorem 7.1].

######  Theorem 2.3.

Let $E/K$ be an elliptic curve. The following are equivalent:

  * •

$E$ has good reduction over $K$;

  * •

$\rho_{E,\ell}$ is unramified for some (all) primes $\ell\neq p$.

So, if $E/K$ has good reduction, the Galois representation factors through the
quotient of $G_{K}$ by $I_{K}$, which is generated by Frobenius.

${G_{K}}$${\operatorname{Aut}(T_{\ell}(E))}$${G_{K}/I_{K}\cong\langle\operatorname{Frob}\rangle}$$\scriptstyle{\rho_{E,\ell}}$

In this case the problem reduces to the study of the image of Frobenius. As it
is shown in [S98, IV §2.3], it is always diagonalisable (at least in an
algebraic closure of $\mathbb{Q}_{\ell}$), so it is enough to compute its
eigenvalues to uniquely determine its action.

######  Lemma 2.4.

The characteristic polynomial of $\rho_{E,\ell}(\operatorname{Frob})$ is
independent of $\ell$ (as explained in [ST68, Theorem 2]) and given by

$\displaystyle F(T)=T^{2}-aT+q,$

where $q=|k|$ and $a=q+1-|\tilde{E}(k)|$, where by $\tilde{E}$ we denote the
reduction of the curve $E$ to $k$.

###### Proof.

See [S98, IV §1.3]. ∎

######  Example 2.5.

Let $K$ be a local field with residue characteristic $3$ and let $k$ be its
residue field. Let $E/K$ be an elliptic curve such that the reduction over $k$
is: $y^{2}=x^{3}-x$. In particular, $E$ has good reduction over $K$.

We want to compute the eigenvalues of $\rho_{E,\ell}(\operatorname{Frob})$,
for a prime $\ell\neq 3$, as elements of $\mathbb{C}$.

First we assume that $k=\mathbb{F}_{3}$. Then the reduced curve $\tilde{E}$
has the following $4$ points over $\mathbb{F}_{3}$:
$\\{O,(0,0),(1,0),(2,0)\\}$, therefore in the notation above $a=0$, $q=3$ and
the roots of $F(T)$, hence the eigenvalues of
$\rho_{E,\ell}(\operatorname{Frob})$, are $\pm i\sqrt{3}$.

If $[k:\mathbb{F}_{3}]=n\geq 1$, then $\rho_{E,\ell}(\operatorname{Frob})$
acts as the $n$-th power of the linear operator described above, so its
eigenvalues are $(\pm i\sqrt{3})^{n}$. In particular we have:

$n\equiv 0\pmod{4}$ | $+3^{n/2}$ | $+3^{n/2}$
---|---|---
$n\equiv 2\pmod{4}$ | $-3^{n/2}$ | $-3^{n/2}$
$n$ odd | $i^{n}\sqrt{3^{n}}$ | $-i^{n}\sqrt{3^{n}}$

Now let us consider elliptic curves with additive, potentially good reduction.
Our first step is to determine the image of the inertia subgroup. As a
consequence of the Criterion of Néron-Ogg-Shafarevich, if the curve $E/K$ has
potentially good reduction then the inertia subgroup $I_{K}$ acts through a
finite quotient on $T_{\ell}(E)$ for some (all) $\ell\neq p$ prime. (See [S86,
VII §7 Corollary 7.3].)

Since we are only interested in the action of
$I_{K}=\operatorname{Gal}(\overline{K}/K^{nr})$ and since the reduction types
of $E$ over $K$ and $K^{nr}$ are the same, we may work simply on $K^{nr}$.
Suppose $E$ has potentially good reduction (recall this is equivalent to the
$j$-invariant of $E$ being integral) and let $L$ be the minimal finite
extension of $K^{nr}$ over which $E$ acquires good reduction. Then it follows
from Theorem 2.3 that for $\ell\neq p$ the Galois representation
$\rho_{E,\ell}$ factors through the quotient $I_{K}/I_{L}$, which is
isomorphic to $\operatorname{Gal}(L/K^{nr})$. Indeed, the subgroup $I_{L}$ is
normal in $I_{K}$, so $L/K^{nr}$ is Galois. This group is also finite, and
since we chose $L$ to be minimal, it injects into
$\operatorname{Aut}(T_{\ell}(E))$.

${\operatorname{Gal}(\overline{K}/K^{nr})}$${\operatorname{Aut}(T_{\ell}(E))}$${\operatorname{Gal}(L/K^{nr})}$$\scriptstyle{\rho_{E,\ell}}$

Furthermore, there is an injection of $\operatorname{Gal}(L/K^{nr})$ into
$\operatorname{Aut}(\tilde{E_{L}})$, where $\tilde{E_{L}}$ is the reduction of
a minimal equation for $E$ over $L$. For more details see [ST68, proof of
Theorem 2]. In particular, the image of inertia does not depend on $\ell$, and
we can restrict the set of possible inertia groups to

$\displaystyle C_{2},C_{3},C_{4},C_{6},$ $\displaystyle C_{3}\rtimes$
$\displaystyle C_{4}\text{ only when }p=3,$ $\displaystyle Q_{8},SL_{2}$
$\displaystyle(\mathbb{F}_{3})\text{ only when }p=2.$

This list comes from the following classification of the automorphisms of an
elliptic curve defined over a field of characteristic $p$ (see [S86, III §10
proof of Theorem 10.1 and Appendix A Proposition 1.2(c), Exercise A.1]).

| $j\neq 0,1728$ | $j=1728$ | $j=0$
---|---|---|---
$p\neq 2,3$ | $C_{2}$ | $C_{4}$ | $C_{6}$
$p=3$ | $C_{2}$ | $C_{3}\rtimes C_{4}$ | $C_{3}\rtimes C_{4}$
$p=2$ | $C_{2}$ | $SL_{2}(\mathbb{F}_{3})$ | $SL_{2}(\mathbb{F}_{3})$

Since these groups have all different orders, if we know the degree of the
extension $L/K^{nr}$, we can uniquely determine the Galois group of this
extension, hence the structure group of the image of inertia. From this moment
on, we denote by $\Phi$ the group $\operatorname{Gal}(L/K^{nr})$. In [K90],
there are complete classification theorems that depend on the residue
characteristic being $2$, $3$ or higher. In this work we focus only on the
case $p=3$. The main lemma to compute the extension $L$ is the following.

######  Lemma 2.6.

With the notations as above, we have

$\displaystyle L=K^{nr}(E[2],\Delta^{1/4}),$

where $\Delta^{1/4}$ is any fourth root of $\Delta$.

###### Proof.

See the Corollary to [K90, Lemma 3]. ∎

Using this result and Tate’s Algorithm (see [S94, IV §9]), Kraus proves the
following classification theorem.

######  Theorem 2.7.

For $p=3$ we have the following classification of the possible images of
inertia:

  * •

if $E$ has type $I^{*}_{0}$, then $v(\Delta)=6$ and $\Phi\cong C_{2}$;

  * •

if $E$ has type $III$, then $v(\Delta)=3$ and $\Phi\cong C_{4}$;

  * •

if $E$ has type $III^{*}$, then $v(\Delta)=9$ and $\Phi\cong C_{4}$;

  * •

if $v(\Delta)\equiv 0\pmod{4}$, then $\Phi\cong C_{3}$;

  * •

if $v(\Delta)\equiv 2\pmod{4}$ and $E$ has type different from $I^{*}_{0}$,
then $\Phi\cong C_{6}$;

  * •

if $v(\Delta)$ is odd and $E$ has type different from $III$ and $III^{*}$,
then $\Phi\cong C_{3}\rtimes C_{4}$.

We refer to [K90, Theorems 2 and 3] for the classification theorem in the case
$p=2$. It is worth mentioning a more general result, which can be found in
[ST68, §2 Corollary 3].

######  Theorem 2.8.

We have $L=K^{nr}(E[m])$, where $m$ is any integer with $m\geq 3$ and
$(p,m)=1$. The Galois group $\operatorname{Gal}(\overline{K}/L)$ is equal to
$\ker(\rho_{E,\ell})$ for any $\ell\neq p$.

##  3\. Curves over a $3$-adic field with non-cyclic inertia action

In the following, we keep the notation and assumptions from the previous
sections. When $p\geq 5$, the extension $L/K^{nr}$ is tamely ramified. Hence,
for all $\ell\neq p$ prime, the Galois representation $\rho_{E,\ell}$ can be
fully described, with the approach in [D16]. A similar approach also works
when $p=3$ but the image of inertia is cyclic and tame. Here we consider the
case when $p=3$ and $E$ has image of inertia isomorphic to $C_{3}\rtimes
C_{4}$, i.e. the last case in Theorem 2.7, and we present an algorithm to
describe completely the Galois action.

###  3.1. Setup

Let $k$ be the residue field of $K$, and let $n=[k:\mathbb{F}_{3}]$. Then $n$
is even if and only if $K$ contains a primitive fourth root of unity, which we
denote $\zeta_{4}$ (see [S95, XIV §3 Lemma 1]). The Galois representation
changes substantially in these two cases. Let us fix some notation. By Lemma
2.6, we have that the minimal field extension $L/K^{nr}$ where $E$ acquires
good reduction is generated by $2$-torsion and a $4$-th root of the
discriminant. Let $F$ be the field extension of $K$ generated by these
elements. Note that since $\operatorname{char}(K)=0$, the curve $E$ can be
written with a Weierstrass equation of the form $y^{2}=f(x)$, where $f$ is a
monic polynomial of degree $3$. If $\alpha,\beta,\gamma$ are the roots of $f$
in $\overline{K}$, $\Delta$ the discriminant of $E$, then
$F=K(\alpha,\beta,\gamma,\Delta^{1/4})$, for some choice of $\Delta^{1/4}$.
Then the discriminant $\Delta_{f}$ of the defining polynomial $f$, which is
given by

$\displaystyle\Delta_{f}=(\beta-\alpha)^{2}(\gamma-\beta)^{2}(\alpha-\gamma)^{2},$

differs from $\Delta$ only by a factor $16$; in particular we have that
$\sqrt{\Delta_{f}}$ and therefore $\sqrt{\Delta}$ are in the field generated
by $2$-torsion.

######  Remark 3.1.

The extension $F/K$ is totally ramified of degree $12$. Indeed, since
$L=FK^{nr}$, $L/F$ is unramified and $L/K^{nr}$ is totally ramified of degree
$12$, we have that $F/K$ has a subextension which is totally ramified of
degree $12$. On the other hand, we have $[F:K]\mid 12$, since the extension of
$K$ generated by $2$-torsion has degree dividing $6$ and it contains a square
root of the discriminant, so $F$ is at most a quadratic extension of it.
Therefore the degree is equal to $12$ and the whole extension $F/K$ is totally
ramified. Moreover, as it has good reduction over $L=FK^{nr}$, $E$ acquires
good reduction over $F$.

However $F/K$ is not necessarily Galois. In fact it is Galois, with Galois
group isomorphic to $\Phi\cong C_{3}\rtimes C_{4}$, if and only if
$\zeta_{4}\in K$, i.e. if $n$ is even. Otherwise, its Galois closure is
$F(\zeta_{4})$ and $\operatorname{Gal}(F(\zeta_{4})/K)$ is isomorphic to the
semidirect product $(C_{3}\rtimes C_{4})\rtimes C_{2}$. As follows from the
classification in [GN], this group is $C_{3}\rtimes D_{4}$. We will now fix a
presentation for the group $\operatorname{Gal}(F(\zeta_{4})/K)$ in the two
cases and show that this group is isomorphic to $C_{3}\rtimes C_{4}$ for $n$
even, $C_{3}\rtimes D_{4}$ for $n$ odd.

Suppose first that $n$ is even. With the notation used above, we define
$\sigma$ and $\tau$ to be the generators of $\operatorname{Gal}(F/K)$ that act
on $\alpha,\beta,\gamma$ and $\Delta^{1/4}$ as follows:

$\displaystyle\sigma:$
$\displaystyle\alpha\mapsto\beta,\quad\beta\mapsto\gamma,\quad\gamma\mapsto\alpha,\qquad\Delta^{1/4}\mapsto\Delta^{1/4};$
$\displaystyle\tau:$
$\displaystyle\alpha\mapsto\alpha,\quad\beta\mapsto\gamma,\quad\gamma\mapsto\beta,\qquad\Delta^{1/4}\mapsto\zeta_{4}\Delta^{1/4}.$

Then $\operatorname{Gal}(F/K)$ has the following presentation

$\displaystyle\langle\sigma,\tau;\sigma^{3}=\tau^{4}=1,\tau\sigma\tau^{-1}=\sigma^{-1}\rangle.$

This group is isomorphic to $C_{3}\rtimes C_{4}$, which is given by the
presentation in [GN], via the isomorphism from $C_{3}\rtimes C_{4}$ to
$\operatorname{Gal}(F/K)$ defined by $a\mapsto\sigma\tau^{2}$ and
$b\mapsto\tau$; in fact it is easy to check that these elements satisfy
$a^{6}=1,b^{2}=a^{3},bab^{-1}=a^{-1}$.

Suppose now that $n$ is odd, i.e. $\zeta_{4}\notin K$. Then the Galois closure
of $F/K$ is given by $F(\zeta_{4})$ and the subgroup generated by the elements
$\sigma,\tau$ is the inertia subgroup of $\operatorname{Gal}(F(\zeta_{4})/K)$.
Furthermore, $\operatorname{Gal}(F(\zeta_{4})/K)$ contains an extra unramified
automorphism corresponding to the map $\zeta_{4}\mapsto-\zeta_{4}$, which we
call $\phi$. Therefore it is presented by the following relations:

$\displaystyle\sigma^{3}=1;\;\tau^{4}=1;\;\phi^{2}=1;\;\sigma\phi=\phi\sigma;\;\phi\tau\phi=\tau^{-1};\;\tau\sigma\tau^{-1}=\sigma^{2}.$

Now $C_{3}\rtimes D_{4}=\langle
a,b,c|a^{3}=b^{4}=c^{2}=1,bab^{-1}=cac=a^{-1},cbc=b^{-1}\rangle$ (see [GN]).
The map from $C_{3}\rtimes D_{4}$ to $\operatorname{Gal}(F(\zeta_{4})/K)$
given by $a\mapsto\sigma$, $b\mapsto\tau$ and $c\mapsto\tau\phi$ is an
isomorphism.

We denote the conjugacy classes of the group obtained in the two cases as
follows (for the sake of completeness, we will use both the notation of [GN]
and the one introduced in this paper):

  * •

if $n$ is even, the conjugacy classes of $\operatorname{Gal}(F/K)\cong
C_{3}\rtimes C_{4}$ are $1=[e],\ 2=[\tau^{2}=b^{2}],\ 3=[\sigma=ab^{2}],\
4A=[\tau=b],\ 4B=[\sigma\tau=ab^{-1}],\ 6=[\sigma\tau^{2}=a]$;

  * •

if $n$ is odd, the conjugacy classes of
$\operatorname{Gal}(F(\zeta_{4})/K)\cong C_{3}\rtimes D_{4}$ are $1=[e],\
2A=[\tau^{2}=b^{2}],\ 2B=[\phi=b^{-1}c],\ 2C=[\tau\phi=c],\ 3=[\sigma=a],\
4=[\tau=b],\ 6A=[\sigma\phi=ab^{-1}c],\ 6B=[\sigma^{2}\phi=a^{2}b^{-1}c],\
6C=[\sigma\tau^{2}=ab^{2}]$.

We can now prove Theorem 1.1.

###  3.2. The case $n$ is even

###### Proof of Theorem 1.1.

Let us first consider the case $n$ even, i.e. $\zeta_{4}\in K$. Then, if $F$
is as at the beginning of Section 3, we showed that $F/K$ is Galois with
Galois group $C_{3}\rtimes C_{4}$. But then $F^{nr}/K$ is the compositum of
the Galois extensions $F/K$ and $K^{nr}/K$, which intersect in $K$ since $F/K$
is totally ramified and $K^{nr}/K$ is unramified. So
$\operatorname{Gal}(F^{nr}/K)\cong\operatorname{Gal}(F/K)\times\operatorname{Gal}(K^{nr}/K)$.
In particular the Frobenius element, that generates
$\operatorname{Gal}(K^{nr}/K)$, commutes with every element of this group,
therefore its image under $\rho$ can be represented as a scalar matrix. We
computed in Example 2.5 the eigenvalues of the Frobenius element of $F$, which
coincide and are equal to $(-3)^{n/2}$ for every even $n$. As $F/K$ is totally
ramified, the image of the Frobenius element of $K$ under $\rho$ is conjugate,
so it has the same eigenvalues.

Now define the following unramified character:

$\displaystyle\chi(\operatorname{Frob})$ $\displaystyle=(-3)^{n/2};$
$\displaystyle\chi\big{|}_{I_{K}}$ $\displaystyle=1.$

Then $\rho(\operatorname{Frob})=\chi(\operatorname{Frob})Id$ and there exists
a representation $\psi$ such that $\rho=\chi\otimes\psi$. The representation
$\psi$ is irreducible of dimension $2$, since $\rho$ is, it is trivial on
Frobenius and coincides with $\rho$ on inertia; therefore it factors through
$C_{3}\rtimes C_{4}$ and, as a representation of this finite group, it is
faithful. The group $C_{3}\rtimes C_{4}$ has only one irreducible faithful
$2$-dimensional representation (see [GN]), so the Galois representation is
completely described by it. Namely the character of $\psi$ in this case is:

$\begin{array}[]{c|cccccc}\rm class&\rm 1&\rm 2&\rm 3&\rm 4A&\rm 4B&\rm
6\cr\rm size&1&1&2&3&3&2\cr\hline\cr\psi&2&-2&-1&0&0&1\cr\end{array}$

as claimed. ∎

###  3.3. The case $n$ is odd

######  Lemma 3.2.

Let $F(\zeta_{4})$ be the Galois closure of $F$. Then:

$\displaystyle
F(\zeta_{4})=K(\sqrt{\beta-\alpha},\sqrt{\gamma-\beta},\sqrt{\alpha-\gamma},\sqrt{\alpha-\beta},\sqrt{\beta-\gamma},\sqrt{\gamma-\alpha}).$

###### Proof.

Let

$\displaystyle
F^{\prime}=K(\sqrt{\beta-\alpha},\sqrt{\gamma-\beta},\sqrt{\alpha-\gamma},\sqrt{\alpha-\beta},\sqrt{\beta-\gamma},\sqrt{\gamma-\alpha}).$

First of all, we prove that $F(\zeta_{4})\subseteq F^{\prime}$. Indeed
$\alpha,\beta,\gamma$ are clearly in $F^{\prime}$;
$\dfrac{\sqrt{\beta-\alpha}}{\sqrt{\alpha-\beta}}$ is a primitive fourth root
of unity contained in $F^{\prime}$; finally one possible choice for
$\Delta^{1/4}$ is given by the product
$2\sqrt{(\beta-\alpha)(\gamma-\beta)(\alpha-\gamma)}$, which is in
$F^{\prime}$. Since $F^{\prime}$ contains this element and a primitive fourth
root of unity, then also all the other fourth roots of $\Delta$ (which are in
$F(\zeta_{4})$) are in $F^{\prime}$. To prove that $F^{\prime}\subseteq
F(\zeta_{4})$, let $B=K(\alpha,\beta,\gamma)$; we show that
$[F^{\prime}:B]\mid[F(\zeta_{4}):B]$. The extension $F(\zeta_{4})/B$ is of
degree $4$, with an unramified subextension of degree $2$ and a totally tamely
ramified subextension of degree $2$. Therefore
$\operatorname{Gal}(F(\zeta_{4})/B)\cong C_{2}\times C_{2}$. The extension
$F^{\prime}/B$ is the compositum of some quadratic extensions, so it is
abelian of exponent $2$. By [S95, XIV §4 Exercise 3], we have
$|B^{\times}/(B^{\times})^{2}|=4$, hence $B^{\times}/(B^{\times})^{2}\cong
C_{2}\times C_{2}$, and by Kummer theory, the abelian extensions of $B$ of
exponent $2$ are in bijection with the subgroups of
$B^{\times}/(B^{\times})^{2}$, which are five, namely $B$, three quadratic
extensions and the biquadratic; therefore $[F^{\prime}:B]\mid 4$. ∎

We now want to compute the action of $\phi$ on the generators of $F^{\prime}$;
for any choices of the square roots, we know that
$\phi(\sqrt{\beta-\alpha})=\pm\sqrt{\beta-\alpha}$ and similarly
$\phi(\sqrt{\alpha-\beta})=\pm\sqrt{\alpha-\beta}$. On the other hand, $\phi$
changes the sign of $\dfrac{\sqrt{\alpha-\beta}}{\sqrt{\beta-\alpha}}$ for it
is a primitive fourth root of unity. Therefore, we have either:

  * •

$\phi(\sqrt{\beta-\alpha})=\sqrt{\beta-\alpha}$ and
$\phi(\sqrt{\alpha-\beta})=-\sqrt{\alpha-\beta}$, or

  * •

$\phi(\sqrt{\beta-\alpha})=-\sqrt{\beta-\alpha}$ and
$\phi(\sqrt{\alpha-\beta})=\sqrt{\alpha-\beta}$.

Without loss of generality the first condition holds, so
$\sqrt{\beta-\alpha}\in F$. Similarly, using the relations between the
generators of $\operatorname{Gal}(F^{\prime}/K)$, we have that $\phi$ fixes
$\sqrt{\gamma-\beta},\sqrt{\alpha-\gamma}$ and changes sign to the other
generators of $F^{\prime}$; therefore $F$, which is the subfield of
$F^{\prime}$ fixed by $\phi$, satisfies

$F=K(\sqrt{\beta-\alpha},\sqrt{\gamma-\beta},\sqrt{\alpha-\gamma}).$

######  Lemma 3.3.

Let $O_{F}$ be the ring of integers of $F$, with maximal ideal
$\mathfrak{m}_{F}$. Then with the same notation as above, we have

$\displaystyle\dfrac{\sigma(x)}{x}\equiv 1\pmod{\mathfrak{m}_{F}},$

for all $x\in O_{F}\setminus\\{0\\}$.

###### Proof.

As $\sigma$ is in the wild inertia subgroup of
$\operatorname{Gal}(F^{\prime}/K)$, which is equal to the first ramification
group by [S95, IV §2 Corollary 1 to Proposition 7], we have $\sigma(x)\equiv
x\pmod{\mathfrak{m}_{F}^{2}}$. If $x\in O_{F}^{\times}$ (i.e. if $x$ is a
unit), then $\sigma(x)/x\equiv 1\pmod{\mathfrak{m}_{F}^{2}}$, hence modulo
${\mathfrak{m}_{F}}$; if $x=\pi$ is a uniformiser of $O_{F}$ of $F$ then by
[S95, IV §2 Proposition 5] we have $\sigma(x)/x\equiv
1\pmod{\mathfrak{m}_{F}}$. In general $x=\pi^{a}u$ where $a$ is a non-negative
integer and $u\in O_{F}^{\times}$, so
$\sigma(x)/x=(\sigma(\pi)/\pi)^{a}\sigma(u)/u\equiv 1\pmod{\mathfrak{m}_{F}}$.
∎

######  Lemma 3.4.

Let $E$ be as before. Then the reduction of some minimal model for $E/F$ on
the residue field is

$\displaystyle\tilde{E}/k:y^{2}=x^{3}-x.$

###### Proof.

First note that we can write, over $F$, the equation for $E$ as follows:

$\displaystyle y^{2}=(x-\alpha)(x-\beta)(x-\gamma).$

Now, operating the change of variables (well-defined over $F$)

$\displaystyle\left\\{\begin{array}[]{cl}x=&x^{\prime}(\beta-\alpha)+\alpha\\\
y=&y^{\prime}(\sqrt{\beta-\alpha})^{3}\end{array}\right.$

we obtain the new equation

$\displaystyle(y^{\prime})^{2}=x^{\prime}(x^{\prime}-1)(x^{\prime}-\lambda),$

where $\lambda=\dfrac{\gamma-\alpha}{\beta-\alpha}$. Finally, note that
$\alpha-\gamma=\sigma^{2}(\beta-\alpha)$, and since $\sigma$ is an element of
the wild inertia subgroup of $\operatorname{Gal}(F^{\prime}/K)$ then by Lemma
3.3, $\dfrac{\sigma^{2}(\beta-\alpha)}{\beta-\alpha}\equiv 1$ in the residue
field. So the reduction in $k$ of $\lambda$ is the same as the reduction of
$-\dfrac{\sigma^{2}(\beta-\alpha)}{\beta-\alpha}$, i.e. $-1$. With simplified
notation, the reduction of $E$ in $k$ is therefore $y^{2}=x^{3}-x$. ∎

###### Proof of Theorem 1.1.

In Example 2.5 we computed the eigenvalues of the Frobenius element of $F$,
and hence of $K$ (as $F/K$ is totally ramified), acting on $E$, which are $\pm
i^{n}\sqrt{3^{n}}$. In particular, since $n$ is odd, they are complex
conjugate and the trace of Frobenius is $0$.

Let $\chi$ be the following unramified character of $G_{K}$:

$\displaystyle\chi(\operatorname{Frob})$ $\displaystyle=i^{n}\sqrt{3^{n}};$
$\displaystyle\chi_{\big{|}{I_{K}}}$ $\displaystyle=1.$

Then
$\rho(\operatorname{Frob})=\chi(\operatorname{Frob})\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right)$ and
$\rho(\operatorname{Frob}^{2})=\chi(\operatorname{Frob})^{2}Id$. Let $F_{2}$
be the field extension of $K$ fixed by $\operatorname{Frob}^{2}$: then it is
an unramified extension of $K$ of degree $2$, i.e. $F_{2}=K(\zeta_{4})$. Also,
in the notation used above, $F^{\prime}=F(\zeta_{4})=FF_{2}$. So the Galois
group described before $\operatorname{Gal}(F(\zeta_{4})/K)$ is generated by
$\sigma,\tau$ and the class of $\operatorname{Frob}$ modulo
$\operatorname{Frob}^{2}$, which we can identify with $\phi$. Moreover, there
exists an irreducible representation $\psi$ of $G_{K}$ such that

$\displaystyle\rho=\chi\otimes\psi;$

to find it, since $\chi$ is a character, it is sufficient to consider
$\psi(g)=\dfrac{1}{\chi(g)}\rho(g)$. The kernel of this representation $\psi$
is precisely $\operatorname{Gal}(\overline{K}/F(\zeta_{4}))$, so $\psi$
factors through the finite group $\operatorname{Gal}(F(\zeta_{4})/K)$ and it
is indeed an irreducible faithful representation of the finite group
$C_{3}\rtimes D_{4}$.

By looking at the character table of $C_{3}\rtimes D_{4}$ (again, see [GN]) it
follows that there are precisely two irreducible faithful representations of
this group of dimension $2$, and they only differ for the character of the two
conjugacy classes generated by the elements $\sigma\phi$ and $\sigma^{2}\phi$.
To uniquely determine $\psi$ we therefore have to compute the trace of the
element $\psi(\sigma\phi)$, and see whether it is $i\sqrt{3}$ or $-i\sqrt{3}$.

We know from Lemma 3.4 that, over $F$, the equation for $E$ is

$\displaystyle E:y^{2}=(x-\alpha)(x-\beta)(x-\gamma)$

and under the change of variables

$\displaystyle x=x^{\prime}(\beta-\alpha)+\alpha;$ $\displaystyle
y=y^{\prime}(\sqrt{\beta-\alpha})^{3},$

we find the minimal model

$\displaystyle E_{min}:y^{2}=x(x-1)(x-\lambda),$

that reduces to

$\displaystyle\tilde{E}:y^{2}=x^{3}-x$

over the residue field. Now let $f(x,y)=(x^{\prime},y^{\prime})$ be the change
of variables above, $\operatorname{red}$ the reduction map:
$E_{min}(\overline{K})\rightarrow\tilde{E}(\overline{k})$ and
$\operatorname{lift}:\tilde{E}(\overline{k})\rightarrow E_{min}(\overline{K})$
be any section of $\operatorname{red}$. Then we can compute the action of any
Galois automorphism $g$ on the reduced curve $\tilde{E}(\overline{k})$ via the
following composition:

$\displaystyle\operatorname{red}\circ f\circ g\circ
f^{-1}\circ\operatorname{lift}.$

So in particular for $g=\sigma\operatorname{Frob}$ we have (recall
$|k|=3^{n}$):

$\displaystyle\left(\tilde{x},\tilde{y}\right)\xrightarrow{\operatorname{lift}}$
$\displaystyle\left(x,y\right)\xrightarrow{f^{-1}}\left(x(\beta-\alpha)+\alpha,y(\sqrt{\beta-\alpha})^{3}\right)$
$\displaystyle\xrightarrow{\sigma\operatorname{Frob}}$
$\displaystyle\left(\sigma(x)^{3^{n}}\sigma(\beta-\alpha)+\beta,\sigma(y)^{3^{n}}(\sigma(\sqrt{\beta-\alpha}))^{3}\right)$
$\displaystyle\xrightarrow{f}$
$\displaystyle\left(\dfrac{\sigma(x)^{3^{n}}\sigma(\beta-\alpha)+\beta-\alpha}{\beta-\alpha},\sigma(y)^{3^{n}}\dfrac{(\sigma(\sqrt{\beta-\alpha}))^{3}}{(\sqrt{\beta-\alpha})^{3}}\right)$
$\displaystyle=$
$\displaystyle\left(\sigma(x)^{3^{n}}\dfrac{\sigma(\beta-\alpha)}{\beta-\alpha}+1,\sigma(y)^{3^{n}}\dfrac{(\sigma(\sqrt{\beta-\alpha}))^{3}}{(\sqrt{\beta-\alpha})^{3}}\right)\xrightarrow{\operatorname{red}}\left(\tilde{x}^{3^{n}}+1,\tilde{y}^{3^{n}}\right)$

Note that:

  * •

the reductions of $\dfrac{\sigma(\beta-\alpha)}{\beta-\alpha}$ and
$\dfrac{\sigma(\sqrt{\beta-\alpha})}{\sqrt{\beta-\alpha}}$ are $1$ by Lemma
3.3;

  * •

$\operatorname{Frob}$ fixes $F$, therefore it acts trivially on the elements
$\beta-\alpha$ and $\sqrt{\beta-\alpha}$;

  * •

since $\sigma$ is an inertia element, $\sigma(x)\equiv x$ and $\sigma(y)\equiv
y$ in $\overline{k}$.

Now, to compute the trace of $\rho(\sigma\operatorname{Frob})$ we use the
formula

$\displaystyle\operatorname{tr}(\rho(g))=\deg(g)+1-\deg(1-g);$

in our case $\deg(g)=\det\rho(\sigma\operatorname{Frob})=3^{n}$ and
$\deg(1-g)$ is the number of points fixed by $\sigma\operatorname{Frob}$, i.e.
the number of solutions (including the point at infinity) of

(3.4) $\displaystyle\left\\{\begin{array}[]{l}x=x^{3^{n}}+1\\\ y=y^{3^{n}}\\\
y^{2}=x^{3}-x.\end{array}\right.$

Let us first consider the case $n=1$. There are no solutions over
$\overline{k}$ to this system of equations, therefore
$\operatorname{tr}(\rho(\sigma\operatorname{Frob}))=3$. But then

$\displaystyle\operatorname{tr}(\psi(\sigma\phi))=\dfrac{1}{i\sqrt{3}}3=-i\sqrt{3}.$

In general, we know that
$\operatorname{tr}(\psi(\sigma\operatorname{Frob}))=\varepsilon i\sqrt{3}$ for
some $\varepsilon\in\\{\pm 1\\}$, and
$\operatorname{tr}(\rho(\sigma\operatorname{Frob}))=\varepsilon
i\sqrt{3}\chi(\sigma\operatorname{Frob})=\varepsilon
i\sqrt{3}i^{n}\sqrt{3^{n}}=\varepsilon
i^{n+1}\sqrt{3^{n+1}}=\varepsilon(-3)^{(n+1)/2}$. The value of $\varepsilon$
can be determined by solving the system of equations above, but for a general
$n$, it cannot be solved directly. However, the number of solutions is
independent on the curve we use, so it is sufficient to work with a fixed
curve. Consider for example the elliptic curve over $\mathbb{Q}_{3}$

$\displaystyle E:y^{2}=x^{3}+9.$

Since its reduction modulo $3$ is $y^{2}=x^{3}$, the valuation of the
discriminant is $7$ and the $j$-invariant is $0$, this curve has potentially
good reduction, and its Néron type is $IV$, so we are in the last case of
Theorem 2.7. Hence the image of inertia is isomorphic to $C_{3}\rtimes C_{4}$.

Let us fix a basis for $\overline{\mathbb{Q}}_{\ell}^{2}$ (with
$\overline{\mathbb{Q}}_{\ell}$ considered as embedded in $\mathbb{C}$), where
the action of Frobenius is given by the matrix

$\displaystyle\rho(\operatorname{Frob})=\left(\begin{array}[]{cc}i\sqrt{3}&0\\\
0&-i\sqrt{3}\end{array}\right);$

then the image of $\sigma$ is either $\left(\begin{array}[]{cc}\zeta_{3}&0\\\
0&\zeta_{3}^{-1}\end{array}\right)$ or its inverse, where
$\zeta_{3}\in\overline{\mathbb{Q}}_{\ell}$ is the primitive third root of unit
$\dfrac{-1+i\sqrt{3}}{2}$. By the computation done for the case $n=1$, we know
$\operatorname{tr}(\rho(\sigma\operatorname{Frob}))=3$ and a simple check
shows that then $\rho(\sigma)=\left(\begin{array}[]{cc}\zeta_{3}^{-1}&0\\\
0&\zeta_{3}\end{array}\right)$. Now let $K$ be an unramified extension of
$\mathbb{Q}_{3}$ of odd degree $n$, so the residue field $k$ is a degree $n$
extension of $\mathbb{F}_{3}$ and the reduction type and Galois representation
of the curve $E$ base-changed to $K$, restricted to inertia is exactly the
same as above. Then
$\rho(\sigma\operatorname{Frob})=\left(\begin{array}[]{cc}\zeta_{3}^{-1}i^{n}\sqrt{3^{n}}&0\\\
0&-\zeta_{3}i^{n}\sqrt{3^{n}}\end{array}\right)$, with trace
$-(-3)^{(n+1)/2}$. Incidentally, this argument proves the following.

######  Lemma 3.5.

The number of solutions of the system of equations (3.4) above is
$3^{n}+(-3)^{(n+1)/2}$.

###### Proof.

We have $\operatorname{tr}(\rho(\sigma\operatorname{Frob}))=-(-3)^{(n+1)/2}$.
On the other hand, we know

$\displaystyle\operatorname{tr}(\rho(\sigma\operatorname{Frob}))$
$\displaystyle=|k|+1-(1+|\\{\text{solutions to \eqref{sys}}\\}|)=$
$\displaystyle=3^{n}-|\\{\text{solutions to \eqref{sys}}\\}|,$

so the number of solutions to (3.4) is precisely
$3^{n}-\operatorname{tr}(\rho(\sigma\operatorname{Frob}))=3^{n}+(-3)^{(n+1)/2}$.
∎

So with the notation above we have $\varepsilon=-1$, and the character of
$\psi$ is the following:

$\begin{array}[]{c|ccccccccc}\rm class&\rm 1&\rm 2A&\rm 2B&\rm 2C&\rm 3&\rm
4&\rm 6A&\rm 6B&\rm 6C\cr\rm
size&1&1&2&6&2&6&2&2&2\cr\hline\cr\psi&2&-2&0&0&-1&0&-i\sqrt{3}&i\sqrt{3}&1\cr\end{array}$

as claimed. In particular, in the proof we computed the character of an
element of the class $6A$. ∎

######  Remark 3.6.

Throughout this work, we have considered the Galois representation $\rho$ on
the Tate module. It is common to consider instead the representation
$\rho_{\acute{e}t}$ on the étale cohomology of the elliptic curve. The two
representations are dual to each other (see [CS86, Theorem 15.1]), so for each
$g\in G_{K}$,

$\displaystyle\rho_{\acute{e}t}(g)=(\rho(g)^{-1})^{t}.$

In this case, it is a convention to consider the Geometric Frobenius
Automorphism $\operatorname{Frob}^{-1}$ in place of the Arithmetic Frobenius
which we used. Therefore, the eigenvalues of
$\rho_{\acute{e}t}(\operatorname{Frob}^{-1})$ are exactly the same as the
eigenvalues of the Arithmetic Frobenius under $\rho$. In conclusion,
$\rho_{\acute{e}t}$ factors as:

$\displaystyle\rho_{\acute{e}t}=\chi_{\acute{e}t}\otimes\psi_{\acute{e}t},$

where

  * •

$\chi_{\acute{e}t}=\chi^{-1}$ (therefore
$\chi_{\acute{e}t}(\operatorname{Frob}^{-1})=\chi(\operatorname{Frob})$);

  * •

$\psi_{\acute{e}t}$ is the only irreducible representation of $C_{3}\rtimes
D_{4}$ with the following character:

$\begin{array}[]{c|rrrrrrrrr}\rm class&\rm 1&\rm 2A&\rm 2B&\rm 2C&\rm 3&\rm
4&\rm 6A&\rm 6B&\rm 6C\cr\rm
size&1&1&2&6&2&6&2&2&2\cr\hline\cr\psi_{\acute{e}t}&2&-2&0&0&-1&0&i\sqrt{3}&-i\sqrt{3}&1\cr\end{array}$

So $\psi$ and $\psi_{\acute{e}t}$ only differ for the trace of the conjugacy
classes of $\sigma\phi$ and $\sigma^{2}\phi$.

A function computing the Galois representation on the étale cohomology of an
elliptic curve defined over a local field is implemented in MAGMA. The
algorithm for the case presented here has recently been improved, using the
results in Theorem 1.1.

### Acknowledgements

The author would like to thank her supervisor Tim Dokchitser for the useful
conversations and corrections. The work was supported by EPSRC.

## References

  * [1]
  * [GN] T. Dokchitser _GroupNames_ http://groupnames.org.
  * [D16] T. Dokchitser and V. Dokchitser, _Euler factors determine local Weil representations_ , in: Journal für die reine und angewandte Mathematik, (2016) (717) 35–46.
  * [K90] A. Kraus, _Sur le défaut de semi-stabilité des courbes elliptiques à réduction additive_ , in: Manuscripta mathematica, (1990) 69(4) 353–385.
  * [CS86] J.S. Milne, _Abelian Varieties_ , in G. Cornell, J.H. Silverman, _Arithmetic geometry_ , Springer-Verlag (1986).
  * [ST68] J.-P. Serre, J. Tate, _Good reduction of abelian varieties_ , in: Annals of Mathematics, (1968) 88(3) 492–517.
  * [S95] J.-P. Serre, _Local Fields_ , (1995) Springer-Verlag New York.
  * [S98] J.-P. Serre, _Abelian $\ell$-adic representations and elliptic curves_, Research Notes in Mathematics, (1998) Vol 7. Peters.
  * [S86] J.H. Silverman _The Arithmetic of Elliptic Curves_ , (1986) Application of Mathematics, Springer-Verlag New York.
  * [S94] J.H. Silverman, _Advanced topics in the arithmetic of elliptic curves_ , (1994) Graduate texts in mathematics, Springer-Verlag.


# Deep Learning-based RF Fingerprint Authentication with Chaotic Antenna
Arrays
††thanks: This work was funded by U.S. National Science Foundation (NSF) under
grant 2233774.

1st Justin McMillen  Dept. of Electrical Engineering
University of South Florida
Tampa, United States
<EMAIL_ADDRESS>2nd Gokhan Mumcu  Dept. of Electrical Engineering
University of South Florida
Tampa, United States
<EMAIL_ADDRESS>3rd Yasin Yilmaz  Dept. of Electrical Engineering
University of South Florida
Tampa, United States
<EMAIL_ADDRESS>
###### Abstract

Radio frequency (RF) fingerprinting is a tool which allows for authentication
by utilizing distinct and random distortions in a received signal based on
characteristics of the transmitter. We introduce a deep learning-based
authentication method for a novel RF fingerprinting system called Physically
Unclonable Wireless Systems (PUWS). An element of PUWS is based on the concept
of Chaotic Antenna Arrays (CAAs) that can be cost effectively manufactured by
utilizing mask-free laser-enhanced direct print additive manufacturing (LE-
DPAM). In our experiments, using simulation data of 300 CAAs each exhibiting 4
antenna elements, we test 3 different convolutional neural network (CNN)
architectures under different channel conditions and compare their
authentication performance to the current state-of-the-art RF fingerprinting
authentication methods.

###### Index Terms:

Physically Unclonable Wireless Systems, RF fingerprinting, Device
authentication, Deep learning, Additive manufacturing, 3D printing

##  I Introduction

As the number of Internet of Things (IoT) devices and amount of wireless data
communication rapidly increase, so does the threat posed by adversarial
parties trying to exploit the vulnerabilities of wireless systems. Hence, it
is vital to develop more secure methods of authentication and communication
while satisfying the quality and efficiency constraints. With current
technology, security at higher levels in the system (software) usually cannot
protect lower layers (such as spoofing and jamming of hardware). Therefore, to
work towards a more secure system, physical layer security features can and
must complement the upper layer defenses, e.g., multi-factor authentication
through RF fingerprinting. With ever increasing technology available to
attackers and the emergence of much faster computing methods, traditional
encryption techniques will not always be as secure as they currently are [1].
With this in mind, it is imperative to find new, harder-to-crack hardware-
based methods to allow for secure systems.

RF fingerprinting is a promising authentication technique for physical layer
security. The classical RF fingerprinting methods utilize the small amplitude,
phase, and frequency variations that are unique to each device due to the
inevitable randomness during the fabrication of the RF integrated circuits
(ICs) connected to the antenna elements [2, 3]. These signatures, while being
detectable by Machine Learning (ML) algorithms [4], are extremely small due to
tight IC fabrication tolerances. State-of-the-art deep neural networks can
only achieve around 63% accuracy in authenticating 250 devices using these
small signatures [5].

We investigate the concept of _Physically Unclonable Wireless Systems_ (PUWS)
as a new hardware security architecture to augment RF fingerprinting-based
authentication. PUWS can effectively be built by additive manufacturing
approaches where multiple devices within the structure can be randomized to
exhibit enhanced and distinct fingerprints. An important element for PUWS is
chaotic antenna arrays (CAAs) where antenna shapes, locations, and feed
networks are randomized [6]. The prior work [6] on CAA assumed that the user
with CAA has knowledge of the wireless channel, its own phase errors, and with
no spatially varying phase errors. Such knowledge by the device is undesired
and mostly impractical for real life applications. In this paper, we extend
the CAA concept to work _without_ the knowledge of wireless channel or its own
errors during authentication by resorting to deep learning-based detection
algorithms. The mask-free laser-enhanced direct print additive manufacturing
(LE-DPAM) technology investigated by our group can play as the key enabler of
such antenna arrays[7, 8, 9, 10]. Introducing randomness will generate RF
fingerprints based on phase errors at the antenna elements with spatial (i.e.,
$\theta$, $\phi$) variance, which will be shown to greatly benefit physical
layer authentication with RF fingerprint enhancement.

##  II Chaotic Antenna Array Model

For the model, we consider aperture coupled rectangular patch antennas and
introduce randomness into the geometry of the patch and the length of a
meandered feed line section inserted between the coupling aperture and feed
point. The superstrate, substrate, and coupling aperture thicknesses are kept
constant. The unperturbed patch is 14.4 x 12 mm2 and resonates at 5.8 GHz with
12% matching bandwidth, well above ISM bandwidth, allowing resonance shifts by
these randomizations to be tolerable. Corner points
$(x_{i},y_{i}),\,i\,\in\,\\{1,2,3,4\\}$ are randomized according to:

$(x_{i},y_{i})=(x_{i}=r\cos\psi+c_{i}\cos\tau_{i},y_{i}+r\sin\psi+c_{i}\sin\tau_{i})$
(1)

where $r\,\in\,U[0,R]$ and $\psi\,\in\,U[0,2\pi]$, denoting a randomized shift
in the center point of the rectangle with respect to the coupling aperture.
$c_{i}\,\in\,U[0,C]$ and $\tau_{i}\,\in\,U[0,2\pi]$ are used to perturb the
corners of the shifted rectangle to change its shape into a trapezoid. $U$
represents a uniform distribution while $R$ and $C$ set maximum limits for the
randomizations, set to 4mm and 0.5mm, respectively. Figure 1 shows a 16
element CAA that can be formed from the randomized antennas. The CAA technique
is suitable for any frequency band and 5.8 GHz ISM is selected for popularity
and availability of components to form a test bed in near future. Antenna
spacing is half-wavelengths as in traditional practice.

Figure 1: Ansys model of a 4$\times$4 5.8 GHz ISM band CAA with randomized
antenna positions, shapes, and feed line lengths.

Randomizing the antenna shape provides the spatial variance property depicted
in Fig. 2. While feed line length randomization alone creates an enhanced
signature transmitted equally in all directions, like the traditional RF
fingerprint, antenna shape randomization causes the phase error to depend on
the direction of radiation as evidenced by the colorful phase distribution in
Fig.2. Histogram data obtained from 1200 antennas shows that phase error is
uniformly distributed across a [0, $\sim 2\pi$] range.

Figure 2: Phase error w.r.t. a reference antenna in polar coordinates. (Left)
Feed line randomization generates a constant error in all transmission
directions; (Right) Antenna geometry randomization creates $\theta,\phi$
dependent error.

##  III Authentication

The present state-of-the-art in RF fingerprinting-based authentication relies
on deep neural networks [5, 11, 12], as opposed to earlier schemes that used
traditional ML techniques (e.g., SVM, kNN, etc.) [13, 14], statistical
detectors [15], and wavelet transforms. Modern deep convolutional neural
networks (CNNs) can successfully authenticate naturally occurring signatures
in the RF chain in idealized setups with a small number of devices, according
to recent literature [11, 12]. However, [5] recently showed in a sizable study
that even for cutting-edge deep CNNs, _naturally occurring RF fingerprints are
insufficient under realistic circumstances with a large number of devices and
changing channel conditions between training and testing_.

Figure 3: Circuit diagram of CAA. Each chaotic antenna element is sequentially
turned on using switches and has a random and direction-dependent phase error
due to its unique geometry.

###  III-A Simulation Setup

For a study on the feasibility of RF fingerprint authentication on PUWS, we
used simulation data describing the phase variation of 1200 antenna elements
and formed 300 CAAs each with 4 antenna elements. The elevation angle $\theta$
and the azimuth angle $\phi$ from the transmitting CAA to the receiver is
randomly selected within [0°, 75°] and [-180°, 180°], respectively. We
simulated a $f=5$ GHz WiFi environment with Rayleigh multipath fading, in
which people may be walking between the device and the router. Considering a
walking speed $V_{walk}$ of 1 m/sec, the maximum Doppler shift $f_{d}$ is
calculated as 16.67 Hz using the formula $f_{d}=(V_{walk}/c)f$, with the
channel coherence time under the Clarke’s model,
$T_{c}=\sqrt{9/{16\pi}f_{d}^{2}}=0.0254$ sec. With a sampling rate of 1 MHz,
$N=1000$ samples are collected within an authentication sequence of 1 msec.

In each authentication sequence, the 4 antennas in a CAA are turned on
sequentially to transmit a complex pilot signal, as illustrated in Fig. 3. The
circuit diagram in Fig. 3 also shows the chaotic antenna elements with random
and direction-dependent phase errors together with the classical digital phase
shifters. The authenticator receives the in-phase and quadrature (I/Q) samples
through Rayleigh multipath fading channels in addition to additive white
Gaussian noise:

$y_{n}=h_{n}x_{n}+w_{n},$ (2)

where $x_{n}=e^{j\alpha_{n}}$ is the transmitted pilot signal with constant
amplitude and the corresponding phase error $\alpha_{n}$,
$h_{n}\sim\mathcal{N}_{c}(0,\sigma_{h}^{2})$ is the zero-mean complex Gaussian
channel coefficient, and $w_{n}\sim\mathcal{N}_{c}(0,\sigma_{w}^{2})$ is the
additive white complex Gaussian noise. The I and Q samples are the real and
imaginary parts of the received signal. With 4 antenna elements in each array
and collecting I and Q samples of the received signal from each antenna, the
dataset used to authenticate a CAA has a size of $N\times 8$, where $N=1000$
is the number of samples within an authentication session. The CAA phase
errors and the received signals through Rayleigh fading channels were
simulated in Matlab. The signal-to-noise ratio (SNR) is set to 20 dB in the
simulations, i.e., $\sigma_{w}^{2}=0.01$. The CAA data is separated using a
(80-10-10)% split for training, validation, and testing, respectively.

To better understand how CAA affects RF fingerprint authentication, 3
different CNN architectures were selected. A basic CNN consisting of two
convolutional layers each with 64 neurons followed by a single dense layer
forms the baseline performance comparison for the following models. The other
models include VGG-16, a 16 layer neural network [16]; ResNet-50, a 50 layer
network which introduces the concept of residual connections between layers
[17]; Inceptionv3, a deep CNN which utilizes a ’network within a network’
strategy to learn features more deeply [18]; and Exception, an Inception based
model which utilizes residual connections and separable convolutional layers
to improve accuracy [19]. These models were originally intended for image
classification, usually taking $244\times 244\times 3$ size inputs, hence
modifications to the top layers were necessary to work with the received I/Q
samples of size $1000\times 8\times 1$, where the 8 columns correspond to the
I and Q signal samples from the 4 antenna elements.

TABLE I: Test Accuracy Model | Accuracy %
---|---
CNN-3 | 93.3
VGG-16 | 93.5
ResNet-50 | 99.2
Inception | 99.9
Xception | 99.9

###  III-B Results

Table I shows the test classification accuracy for each of the networks
trained and tested in Python using TensorFlow. Even the baseline CNN-3 scores
significantly above the 63% accuracy, which is the existing state-of-the-art
performance in the literature achieved by ResNet-50 using the traditional
(non-CAA) RF fingerprints [5]. The more advanced networks, VGG-16 and
ResNet-50, Inception, and Xception, all achieve much higher accuracies.
Comparing the performances of ResNet-50 using the CAA fingerprints ($>99\%$)
to the traditional RF fingerprints ($63\%$) [5] in similar setups, we see that
the CAA fingerprints enable significantly enhanced authentication capacity
with the help of deep neural networks.

##  IV Conclusion

As dishonest actors develop more advanced security cracking technologies, it
is vital that security keep in-step, lest vital systems become vulnerable to
attack. More than ever there is a need for fast, cheap, secure authentication
methods. In this vein, we investigated a novel authentication concept and
implementation of RF fingerprinting through chaotic antenna arrays (CAAs). By
testing 5 popular convolutional neural network (CNN) architectures, we showed
that deep learning-based authentication utilizing CAA fingerprints
significantly outperforms the existing state-of-the-art results using
traditional RF fingerprints found in all communication devices. Compared to
the 63% accuracy [5] achieved by ResNet-50, a popular CNN architecture, using
the traditional RF fingerprints, the CAA fingerprints enable over 99% accuracy
by ResNet-50 in the task of authenticating 300 devices under Rayleigh fading
channels.

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