# Local Galois representations and Frobenius traces

Tim Dokchitser  Department of Mathematics, University Walk, Bristol BS8 1TW,
UK<EMAIL_ADDRESS>and  Vladimir Dokchitser  University College
London, London WC1H 0AY, UK<EMAIL_ADDRESS>
Ramified $l$-adic local Galois representations tend to be unwelcome. We give
an explicit character formula which reconstructs such a representation from
its Frobenius traces over extensions where it is unramified. We also record
some consequences for étale cohomology of abelian varieties and other $l$-adic
representations. The paper is a number-theoretic counterpart of [12], which
concerns representations of finite groups.

Our setting is as follows. Let $K$ be a non-archimedean local field, with
residue field ${\mathbb{F}}_{K}$, separable closure $K^{s}$ and algebraic
closure $\bar{K}$. We write $I_{K}=I_{K^{s}/K}$ for the absolute inertia group
and $\operatorname{Frob}_{K}\in G_{K}=\operatorname{Gal}(K^{s}/K)$ for an
arithmetic Frobenius element. We refer the reader to [21] for the background
on Weil and Weil-Deligne representations.

We begin with an analogue of the character formula [12, Cor. 9] for Weil
representations:

######  Theorem 1.

Let $F/K$ be a finite Galois extension of local fields and $\rho$ a semisimple
Weil representation that factors through $\operatorname{Gal}(F^{nr}/K)$. Let
$\\{\rho_{i}\\}_{i\in\Lambda}$ be a set of irreducible representations of
$\operatorname{Gal}(F/K)$, with exactly one in each set of unramified
twists111i.e. $\rho_{i}\neq\rho_{j}\otimes$(1-dim) unramified, and each
irreducible of $\operatorname{Gal}(F/K)$ is $\rho_{i}\otimes$(1-dim
unramified) for some $i$. Write $I<\operatorname{Gal}(F/K)$ for the inertia
group and
$m_{i}=\langle\operatorname{Res}_{I}\rho_{i},\operatorname{Res}_{I}\rho_{i}\rangle$.
Then

  * (i)

There are unramified representations $\Psi_{i}$ such that
$\rho\cong\bigoplus_{i\in\Lambda}\rho_{i}\otimes\Psi_{i}.$

  * (ii)

Fix $i\in\Lambda$. For every $d>0$,
$\operatorname{Tr}\Psi_{i}(\operatorname{Frob}_{K}^{dm_{i}})=\frac{1}{|I|\,m_{i}}\sum_{L}\overline{\operatorname{Tr}\rho_{i}(\operatorname{Frob}_{L})}\,\operatorname{Tr}\rho(\operatorname{Frob}_{L}),$
where $L$ ranges over extensions of $K$ in $\smash{F^{nr}}$ of ramification
degree $|I|$ and residue degree $dm_{i}$.

  * (iii)

The twist $\rho_{i}\otimes\Psi_{i}$ is uniquely determined by (ii).

  * (iv)

Concretely, suppose $\dim\Psi_{i}\\!\leq\\!N$. There is a unique
$0\\!\leq\\!n\\!\leq\\!N$ and
$\lambda_{1},...,\lambda_{n}\in\mathbb{C}^{\times}$ such that
$\sum_{k}\lambda_{k}^{d}=\operatorname{Tr}\Psi_{i}(\operatorname{Frob}_{K}^{dm_{i}})$
for $d=1,...,N$. Then $\Psi_{i}(\operatorname{Frob}_{K})$ has eigenvalues
$\sqrt[m_{i}]{\lambda_{1}},...,\sqrt[m_{i}]{\lambda_{n}}$ for some choice of
the roots, and $\rho_{i}\otimes\Psi_{i}$ is independent of this choice.

Let $\ell$ be a prime different from the residue characteristic of $K$. As an
application, we deduce that the $\ell$-adic representation
$\rho_{\ell}=H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{\ell})$ of an
abelian variety $A/K$ is determined by the traces
$\operatorname{Tr}\rho^{I_{L}}_{\ell}(\operatorname{Frob}_{L})$ over the
fields $L$ where $A$ is semistable. (Here, as usual, $\rho_{\ell}^{I_{L}}$ is
the maximal subrepresentation unramified over $L$.) In fact, we can choose a
universal finite list of such fields to control all abelian varieties of a
fixed dimension.

######  Theorem 2.

Let $g\geq 0$. There is a finite Galois extension $K_{g}/K$ such that

  1. (i)

Every $g$-dimensional abelian variety $A/K$ has semistable reduction over
$K_{g}$.

  2. (ii)

The $G_{K}$-representation
$\rho_{\ell}=H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{\ell})$ is uniquely
determined by the traces
$\operatorname{Tr}\rho^{I_{L}}_{\ell}(\operatorname{Frob}_{L})$ over subfields
$K\subset L\subset K_{g}$ for which $K_{g}/L$ is unramified.

  3. (iii)

When $g\geq 2$, for every smooth projective curve $X/K$ of genus $g$ the
$G_{K}$-representation $H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{\ell})$
is uniquely determined by point counts
$\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L})$ with $L$ as in (ii). Here
$\mathcal{X}_{L}$ is the minimal regular model of $X/L$ with special fibre
$\bar{\mathcal{X}}_{L}/{\mathbb{F}}_{L}$.

Theorem 1 allows one to explicitly reconstruct local Galois representations
from the restrictions to subgroups where they become semistable or unramified
(extending [11]). For example, suppose $X/K$ is a curve with very bad
reduction. Then we can identify the Galois representation
$\rho_{l}=H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{\ell})$, provided we
know a Galois extension $F/K$ where $X$ becomes semistable, and we can counts
points on the reduction of $X$ in extensions of $K$ where $X$ is semistable.
We restore the decomposition of $\rho_{l}$ into irreducibles via a weighted
average of Frobenius traces as in Theorem 1 (ii), which can be computed as
point counts on the reduced curves. We give a numerical example at the end of
this paper, which illustrates this in detail. This style of argument has been
used for specific Galois groups (see [11, §§3-4], [10, Thm 7.3], [5], [16,
App. A]), and Theorem 1 gives a universal character formula which applies for
arbitrary groups.

Here are some theoretical consequences for general representations. The
following corollary is essentially due to Saito [18, Lemma 1(1)]; see also
Remark 9.

######  Corollary 3.

Every semisimple Weil representation $\rho$ over $K$ is uniquely determined by
the traces $\operatorname{Tr}\rho(\operatorname{Frob}_{L}),$ where $L$ varies
over finite separable extensions of $K$ over which $\rho$ is unramified.

######  Notation.

Recall that a Frobenius semisimple Weil-Deligne representation
$\rho=(\rho_{\rm Weil},N)$ over $K$ can be decomposed as
$\rho=\bigoplus_{n\geq 1}\rho_{n}\otimes\operatorname{Sp}_{n},$ where
$\rho_{n}$ are semisimple Weil representations (i.e. have $N=0$), and
$\operatorname{Sp}_{n}$ is the $n$-dimensional special representation [21,
4.1.4–4.1.5]. In particular, $\rho^{N=0}:=\ker N=\bigoplus\rho_{n}$.

We say that $\rho$ is _weight-monodromy compatible_ if the eigenvalues of
$\operatorname{Frob}_{K}$ on $\rho_{i}$ are Weil numbers of absolute value
$q^{\frac{n-1}{2}}$, where $q=|{\mathbb{F}}_{K}|$. We use the same term for
$\ell$-adic representations when their associated Weil-Deligne representation
is weight-monodromy compatible.

For every proper smooth variety $X/K$ and $0\leq i\leq 2\dim X$,
$H_{\text{\'{e}t}}^{i}(X_{\bar{K}},\mathbb{Q}_{\ell})(\frac{i}{2})$ is
conjectured to be weight-monodromy compatible [8], and this is often known
(see [19, §1] for a summary). Because a weight-monodromy compatible
representation is determined by $\rho^{N=0}$, from Corollary 3 we get

######  Corollary 4.

Frobenius-semisimple Weil-Deligne representations $\rho=(\rho_{\rm Weil},N)$
over $K$ satisfying the weight-monodromy compatibility are uniquely determined
by the traces $\operatorname{Tr}\rho^{N=0}(\operatorname{Frob}_{L})$, where
$L$ varies over finite separable extensions of $K$ over which $\rho_{\rm
Weil}$ is unramified.

######  Corollary 5.

Let $X/K$ be a proper smooth variety and $0\leq i\leq 2\dim X$. If the
$G_{K}$-representation
$\rho_{\ell}=H_{\text{\'{e}t}}^{i}(X_{\bar{K}},\mathbb{Q}_{\ell})(\frac{i}{2})$
is Frobenius semisimple and weight-monodromy compatible, then it is uniquely
determined by $\operatorname{Tr}\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$
for finite separable extensions $L/K$ for which $I_{L}$ acts unipotently. In
particular, if these traces are independent of $\ell$, then so is the Weil-
Deligne representation associated to
$H_{\text{\'{e}t}}^{i}(X_{\bar{K}},\mathbb{Q}_{\ell})$.

For semistable abelian varieties these traces are known to be independent of
$\ell$ [14, IX, Thm 4.3], and we get the well-known independence of $\ell$ for
Weil-Deligne representations of general abelian varieties (see [6, Ex. 8.10],
[13, Rem. 2.4.6] and [1, Prop. 2.8.1] for a proof):

######  Corollary 6.

For an abelian variety $A/K$, the Weil–Deligne representation associated to
$H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{\ell})$ is independent of
$\ell$.

Finally, we also have a version of Theorem 2 for semisimple $\ell$-adic
representations:

######  Theorem 7.

Let $n\geq 0$ and $\mathcal{F}/\mathbb{Q}_{\ell}$ a finite extension. There is
a finite Galois extension $K_{\mathcal{F},n}/K$ such that all continuous
semisimple representations
$\rho_{\ell}:G_{K}\to\operatorname{GL}_{n}(\mathcal{F})$ are

  1. (1)

unramified over subfields $K\subset L\subset K_{\mathcal{F},n}$ for which
$K_{\mathcal{F},n}/L$ is unramified;

  2. (2)

uniquely determined by the traces
$\operatorname{Tr}\rho_{\ell}(\operatorname{Frob}_{L})$ for $L$ as in (1).

Moreover, any finite extension $E/K_{\mathcal{F},n}$, with $E$ Galois over
$K$, has this property.

Again, because a Frobenius semisimple weight-monodromy compatible
representation is determined by its maximal semisimple subrepresentation, we
deduce

######  Corollary 8.

All Frobenius semisimple weight-monodromy compatible
representations$\rho_{\ell}\colon G_{K}\to\operatorname{GL}_{n}(\mathcal{F})$
are uniquely determined by the traces
$\operatorname{Tr}\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$ with $L$ and
$K_{\mathcal{F},n}$ (or $E$) as in the theorem.

######  Remark 9.

Our strategy of proving that Weil representations are determined by their
Frobenius traces over extensions where they are unramified is known, and goes
back at least to Chebotarev. He used the fact that after adjoining enough
roots of unity, a finite extension of number fields has abundance of
cyclotomic extensions inside it to reduce his density theorem to Hecke’s
cyclotomic case (see e.g. [17]). Similarly, Corollary 3 relies on the abudance
of unramified extensions inside $F^{nr}/K$ for a finite Galois extension
$F/K$. What is new is Theorem 1, which makes this explicit in practice.

As mentioned above, Corollary 3 is due to Saito [18, Lemma 1(1)]. His
assumption $n(\sigma)\geq 0$ is slightly stronger but the proof only uses
$n(\sigma)>0$, which is equivalent to that in Corollary 3. The proof of [18,
Lemma 1(1)] does not explain why $m$ is a function of the traces, and the
field cut out by $\sigma$ is not a finite extension of $K$. However, both of
these issues can be circumvented with an argument similar to the proof of
Corollary 3 below, by comparing two representations instead of taking one.

## Proof of Theorem 1

(i) This is standard; see e.g. [11, Lemma 2.3].

(ii) By linearity in $\rho$, it suffices to prove this when
$\rho=\rho_{i}\otimes\Psi$, with $\Psi$ unramified 1-dimensional. Because the
formula is also invariant under twisting by unramified 1-dimensional
characters, we may assume $\Psi$ factors through $\operatorname{Gal}(F/K)$,
for example $\Psi=\boldsymbol{1}$. This case is proved in [12, Cor 9 (iii)];
note that ranging over $L$ is the same as ranging over
$g\in\operatorname{Frob}_{K}^{dm_{i}}\\!I_{F^{nr}/K}$.

(iii),(iv) This is essentially [12, Cor 9 (iv)]. It is stated there for finite
groups, but the proof is the same: (iii) and (iv) follow formally from (ii)
plus the fact that $\rho_{i}$ is invariant under twisting by unramified
characters of order $m_{i}$, which is [12, Lemma 5]. ∎

## Proof of Corollary 3

Let $\rho_{1},\rho_{2}$ be two non-isomorphic semisimple Weil representations
over $K$. They have finite inertia images, and so there is a finite Galois
extension $F/K$ such that both $\rho_{1}$ and $\rho_{2}$ factor through
$\operatorname{Gal}(F^{nr}/K)$. By Theorem 1, $\rho_{1}$ and $\rho_{2}$ can be
distinguished from one another by their Frobenius traces over extensions $L/K$
inside $F^{nr}$ over which they are unramified.

## Proof of Theorem 7

Fix $n$ and $\mathcal{F}$ as in the theorem. First consider continuous
representations $\tau:G_{K}\to\operatorname{GL}_{n}(\mathcal{F})$ with finite
image. After taking a $\tau(G_{K})$-invariant $O_{\mathcal{F}}$ lattice, we
may assume that $\tau$ lands in $\operatorname{GL}_{n}(O_{\mathcal{F}})$.
Recall that $\operatorname{GL}_{n}(O_{\mathcal{F}})$ has a compact open
subgroup $U$ with no elements of finite order (e.g. using exp and log as in
[20, Appendix]). Therefore
$|\tau(G_{K})|\leq(\operatorname{GL}_{n}(O_{\mathcal{F}}):U)$, and $\tau$
factors thorough a Galois extension of $K$ of at most this degree. There are
only finitely many separable extensions of $K$ of a given degree, and so there
is a finite Galois extension $F/K$, such that every $\tau$ with finite image
factors through $\operatorname{Gal}(F/K)$. It follows that every continuous
semisimple representation $\rho:G_{K}\to\operatorname{GL}_{n}(\mathcal{F})$
factors through $\operatorname{Gal}(F^{nr}/K)$.

Now apply Theorem 1. Let $K_{\mathcal{F},n}$ be the compositum of all $L/K$ in
$F^{nr}$ with ramification degree $e_{L/K}=e_{F/K}$ and residue degree over
$K$ bounded by $n^{3}$. By the theorem (parts (ii)-(iv) and using that
$dm_{i}\leq n^{3}$ in (iv)), Frobenius traces over these fields determine
$\rho_{\mathbb{C}}:G_{K}\to\operatorname{GL}_{n}(\mathbb{C})$ for some chosen
embedding $\mathcal{F}\lhook\joinrel\rightarrow\mathbb{C}$. Therefore $\rho$
itself is uniquely determined as a representation to
$\operatorname{GL}_{n}(\mathcal{F})$. The same is true for any finite
extension $E$ of $K_{\mathcal{F},n}$, Galois over $K$. ∎

## Proof of Theorem 2

We will prove a slightly stronger statement:

######  Theorem 10.

Fix $g,c\in\mathbb{N}$ and a prime $\ell$ different from the residue
characteristic of $K$. There is a finite Galois extension $K_{\ell,g,c}/K$
such that

  1. (i)

Every $g$-dimensional abelian variety $A/K$ has semistable reduction over
$K_{\ell,g,c}$.

  2. (ii)

The $G_{K}$-representation
$\rho_{\ell}=H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{\ell})$ is uniquely
determined by $\operatorname{Tr}\rho^{I_{L}}_{\ell}(\operatorname{Frob}_{L})$
for subfields $K\subset L\subset K_{\ell,g,c}$ with $|{\mathbb{F}}_{L}|>c$ and
$K_{\ell,g,c}/L$ unramified.

  3. (iii)

When $g\geq 2$, for every smooth projective curve $X/K$ of genus $g$, the
$G_{K}$-representation $H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{\ell})$
is uniquely determined by point counts
$\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L})$ with $L$ as in (ii). Here
$\mathcal{X}_{L}$ is the minimal regular model of $X/L$ with special fibre
$\bar{\mathcal{X}}_{L}/{\mathbb{F}}_{L}$.

###### Proof.

(i), (ii) Suppose $c=1$. Recall that
$H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{\ell})$ is Frobenius semisimple
and weight-monodromy compatible [14, IX]. For $L/K$ finite, $I_{L}$ acts
unipotently on it if and only if $A/L$ is semistable [14, IX, 3.5/3.8], [2,
7.4.6]. By Theorem 7 and Corollary 8 for $n=2g$ and
$\mathcal{F}=\mathbb{Q}_{\ell}$, any choice of $K_{\ell,g,1}$ that contains
$K_{\mathcal{F},n}$ satisfies (i) and (ii).

Now let $c$ be arbitrary, and choose a prime $d\neq\ell$ so that
$[\mathbb{Q}_{\ell}(\zeta_{d}):\mathbb{Q}_{\ell}]>4g^{2}$ and
$|{\mathbb{F}}_{K}|^{d}>c$. (All but finitely many primes satisfy these
conditions.) Let $K^{\prime}$ be the unramified extension of $K$ of degree
$d$. Construct $K^{\prime}_{\ell,g,1}$ by the $c=1$ case over $K^{\prime}$,
enlarging it if necessary to contain $K_{\mathcal{F},n}$. We claim that
$K_{\ell,g,c}=K^{\prime}_{\ell,g,1}$ satisfies (i) and (ii). By construction,
(i) holds.

Suppose $A/K$ is an abelian variety. Frobenius traces over subfields
$K^{\prime}\subset L\subset K^{\prime}_{\ell,g,1}$ as in (ii) determine the
restriction of $\rho_{\ell}$ to $K^{\prime}$ uniquely by the properties of
$K_{\mathcal{F},n}$. Such $L$ have
$|{\mathbb{F}}_{L}|\geq|{\mathbb{F}}_{K^{\prime}}|=|{\mathbb{F}}_{K}|^{d}>c$,
so it suffices to show that this restriction determines $\rho$.

By Theorem 7 and Corollary 8 again, it suffices to reconstruct Frobenius
traces over subfields $K\subset L\subset K_{\mathcal{F},n}$ with
$K_{\mathcal{F},n}/L$ unramified. If $L$ contains $K^{\prime}$ then we are
done, as we have already reconstructed the restriction of $\rho_{\ell}$ to
$K^{\prime}$. Suppose $L\not\supset K^{\prime}$, and let
$\alpha_{1},...,\alpha_{m}$ ($m\leq 2g$) be the eigenvalues of
$\rho^{I_{L}}(\operatorname{Frob}_{L})$. As $LK^{\prime}/L$ has degree $d$ (as
$d$ is prime), we know $\alpha_{1}^{d},...,\alpha_{m}^{d}$, the Frobenius
eigenvalues over $LK^{\prime}$ from the restriction to $K^{\prime}$. Now,
$\alpha_{1}^{d}$ has a unique $d$th root that generates an extension of
$\mathbb{Q}_{\ell}$ of degree $\leq 2g$, namely $\alpha_{1}$. Indeed, if
$\alpha_{1}\zeta_{d}^{i}$ also generates such an extension, then
$[\mathbb{Q}_{\ell}(\alpha_{1},\zeta_{d}):\mathbb{Q}_{\ell}]\leq 4g^{2}$. But
$[\mathbb{Q}_{\ell}(\zeta_{d}):\mathbb{Q}_{\ell}]>4g^{2}$ by the choice of
$d$. Therefore $\alpha_{1},...,\alpha_{m}$ are indeed determined by
$\alpha_{1}^{d},...,\alpha_{m}^{d}$, as required.

(iii) We claim that any $K_{\ell,g,1}$ that satisfies (i) and (ii) also
satisfies (iii), provided $c>16g^{2}$ (which we may clearly assume).

Let $A$ be the Jacobian of $X$. Recall that there is a canonical isomorphism

$H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{l})\cong
H_{\text{\'{e}t}}^{1}(A_{\bar{K}},\mathbb{Q}_{l}).$

By (ii), this representation, say $\rho_{\ell}$, is determined by the traces
$\operatorname{Tr}\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$ for $L$ as in
(ii). By (i), over these fields $A$ is semistable, hence so is $X$ as $g\geq
2$ [9, Thm. 1.2]. Therefore

$H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{l})^{I_{L}}\cong
H_{\text{\'{e}t}}^{1}((\bar{\mathcal{X}}_{L})_{\bar{\mathbb{F}}_{L}},\mathbb{Q}_{l})$

as $\operatorname{Gal}(\bar{\mathbb{F}}_{L}/{\mathbb{F}}_{L})$-modules, see
e.g. [10, Thm. B.1]. By the Grothendieck-Lefschetz trace formula [7, p. 86,
Thm 3.1],

$\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L})=t_{0}-t_{1}+t_{2},$

where $t_{i}$ is the trace of the Frobenius automorphism on
$H_{\text{\'{e}t}}^{i}((\bar{\mathcal{X}}_{L})_{\bar{\mathbb{F}}_{L}},\mathbb{Q}_{l})$.
In particular,
$t_{1}=\operatorname{Tr}\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$ by the
above. Note that $t_{0}=1$ as $\bar{\mathcal{X}}_{L}$ is connected, and
$t_{2}$ is a multiple of $q=|{\mathbb{F}}_{L}|$ as
$H_{\text{\'{e}t}}^{2}((\bar{\mathcal{X}}_{L})_{\bar{\mathbb{F}}_{L}},\mathbb{Q}_{l})$
is the permutation module on the irreducible components of
$\bar{\mathcal{X}}_{L}$ twisted by 1. Now, $|t_{1}|\leq 2g\sqrt{q}$ because
all eigenvalues of $\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$ have
absolute values 1 or $\sqrt{q}$ and $\dim\rho_{\ell}^{I_{L}}\leq 2g$. Because
$q\geq c>16g^{2}$, we have $2g\sqrt{q}<q/2$, and so

$t_{1}\bmod
q\>\>\>=\>\>\>(t_{0}+t_{2}-\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L}))\bmod
q\>\>\>=\>\>\>(1-\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L}))\bmod q$

recovers $t_{1}$ uniquely. This shows that
$\\#\bar{\mathcal{X}}_{L}({\mathbb{F}}_{L})$ determines
$t_{1}=\operatorname{Tr}\rho_{\ell}^{I_{L}}(\operatorname{Frob}_{L})$, and the
collection of these for varying $L$ determines
$H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{l})$. ∎

## An example

We end with a numerical example of how Theorem 1 works in practice.

Let $K=\mathbb{Q}_{7}$ and let $F/\mathbb{Q}_{7}$ be a wildly ramified Galois
extension of degree 21 with (non-abelian) Galois group $G=C_{7}\rtimes C_{3}$,
residue degree 3 and ramification degree 7. The character table of $G$ is as
follows, after fixing a choice of a cube root of unity $\zeta_{3}$ and
$\sqrt{-7}$ in $\mathbb{C}$:

$\begin{array}[]{c|ccccc}&\rm
1&\tau&\tau^{-1}&\sigma&\sigma^{-1}\cr\hline\cr\boldsymbol{1}&1&1&1&1&1\cr\chi&1&\zeta_{3}^{2}&\zeta_{3}&1&1\cr\bar{\chi}&1&\zeta_{3}&\zeta_{3}^{2}&1&1\cr\rho_{1}&3&0&0&\frac{-1-\sqrt{-7}}{2}&\frac{-1+\sqrt{-7}}{2}\cr\rho_{2}&3&0&0&\frac{-1+\sqrt{-7}}{2}&\frac{-1-\sqrt{-7}}{2}\cr\end{array}$
Table 1. Character table of $G=C_{7}{\rtimes}C_{3}$

Consider a genus 3 curve $X$ whose Jacobian has bad reduction over
$\mathbb{Q}_{7}$, but $X$ acquires good reduction over $F$. Our goal is to
determine the Galois representation
$\rho=H_{\text{\'{e}t}}^{1}(X_{\bar{K}},\mathbb{Q}_{\ell})\otimes_{\mathbb{Q}_{\ell}}\mathbb{C}$.
By the Néron-Ogg-Shafarevich criterion, $I_{K}$ acts on $V$ through $C_{7}$.
By Theorem 1(i), and the fact that the characteristic polynomials of inertia
elements have rational coefficients,

$\rho\cong(\rho_{1}\otimes\Psi_{1})\oplus(\rho_{2}\otimes\Psi_{2}),$

for some unramified characters $\Psi_{1},\Psi_{2}$.

Theorem 1 lets us determine the $\Psi_{i}$ as follows. Let $\tilde{F}$ be the
degree 7 unramified extension of $F$. For each element $g\in
I_{F/\mathbb{Q}_{7}}=C_{7}$ let
$\tilde{g}\in\operatorname{Gal}(\tilde{F}/\mathbb{Q}_{7})$ be the unique
element that projects to $g\in\operatorname{Gal}(F/\mathbb{Q}_{7})$ and acts
as $\operatorname{Frob}_{\mathbb{Q}_{7}}^{3}$ on the residue field. Write
$F_{g}=F^{\langle\tilde{g}\rangle}$ for the corresponding degree 21 extensions
of $\mathbb{Q}_{7}$. Note that $\operatorname{Jac}X$ has good reduction over
$F_{g}$, since $\tilde{F}/F_{g}$ is unramified.

By the theorem,

$\Psi_{i}(\operatorname{Frob}_{\mathbb{Q}_{7}}^{3})=\frac{1}{|C_{7}|}\frac{1}{\langle\operatorname{Res}_{C_{7}}\rho_{i},\operatorname{Res}_{C_{7}}\rho_{i}\rangle}\sum_{g\in
I_{F/\mathbb{Q}_{7}}}\overline{\operatorname{Tr}\rho_{i}(g)}\operatorname{Tr}\rho(\operatorname{Frob}_{F_{g}})=$
$=\frac{1}{21}\sum_{g\in
I_{F/\mathbb{Q}_{7}}}\overline{\operatorname{Tr}\rho_{i}(g)}\cdot(7^{3}+1-\\#\overline{\mathcal{X}_{F_{g}}}(\mathbb{F}_{7^{3}})),$

where $\mathcal{X}_{F_{g}}$ is a regular model of $X$ over $F_{g}$, by the
Grothendieck-Lefschetz trace formula (as in proof of Theorem 10 (iii)).
Moreover, $\Psi_{i}(\operatorname{Frob}_{\mathbb{Q}_{7}})$ can be taken to be
any cube root of this value, as the discrepancy vanishes when taking the
tensor product with $\rho_{i}$.

As a particular example, let $F/\mathbb{Q}_{7}$ be the splitting field of

$f(x)=x^{7}+21x^{6}+7,$

taken from [15]. Let $\alpha$ be a root of $f$ in $F$; it is a uniformiser of
$F$. Let $\zeta\in F$ be a primitive 18th root of unity. One checks that the
roots $\alpha_{j}$ of $f$ have expansions

$\alpha_{j}=\alpha+j\zeta\alpha^{2}+O(\alpha^{3}),\qquad j=0,...,6.$

The generators of $G$ are then determined by

$\sigma\colon\alpha_{j}\mapsto\alpha_{j+1}\text{ (order
7)},\qquad\tau\colon\alpha_{j}\mapsto\alpha_{2j}\text{ (order 3)},$

with indices taken modulo 7.

Now take $X:y^{2}=f(x)$. It acquires good reduction over
$\mathbb{Q}_{7}(\alpha)$, and using Magma [3] we find that

$\\#\overline{\mathcal{X}}_{F_{\operatorname{id}}}(\mathbb{F}_{7^{3}})=7^{3}+1=344,$
$\\#\overline{\mathcal{X}}_{F_{\sigma}}(\mathbb{F}_{7^{3}})=\\#\overline{\mathcal{X}}_{F_{\sigma^{2}}}(\mathbb{F}_{7^{3}})=\\#\overline{\mathcal{X}}_{F_{\sigma^{4}}}(\mathbb{F}_{7^{3}})=7^{3}-7^{2}+1=295,$
$\\#\overline{\mathcal{X}}_{F_{\sigma^{3}}}(\mathbb{F}_{7^{3}})=\\#\overline{\mathcal{X}}_{F_{\sigma^{5}}}(\mathbb{F}_{7^{3}})=\\#\overline{\mathcal{X}}_{F_{\sigma^{6}}}(\mathbb{F}_{7^{3}})=7^{3}+7^{2}+1=393,$

which leads to $\Psi_{1}(\operatorname{Frob}_{\mathbb{Q}_{7}})=\sqrt{-7}$ and
$\Psi_{2}(\operatorname{Frob}_{\mathbb{Q}_{7}})=-\sqrt{-7}$.

Conversely, let $X^{\prime}:y^{2}=x^{7}+420x^{6}-245x^{3}+1225x^{2}-833x+189$.
(The right-hand side is the minimal polynomial of $-\alpha^{2}-\alpha$.) This
curves acquires good reduction over the same fields as $X$, but has the
opposite point counts (that is, 295 and 393 are interchanged), so that here
$\Psi_{1}(\operatorname{Frob}_{\mathbb{Q}_{7}})=-\sqrt{-7}$ and
$\Psi_{2}(\operatorname{Frob}_{\mathbb{Q}_{7}})=\sqrt{-7}$.

What makes the example interesting is that the two curves have exactly the
same point counts over all extensions of $\mathbb{Q}_{7}$ of degree at most
20. In that sense our result here is best possible.

### Acknowledgements

We would like to thank Toby Gee and Aurel Page for helpful discussions. The
second author is supported by a Royal Society University Research Fellowship.

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