Here K will denote a p-adic field and F and `-adic field, where p and ` are two distinct
primes.
ΦK = geometric Frobenius
IK = absolute inertia group of K
w = absolute wild inertia group of K
IK
Theorem 1. Let τ : GK → GLd (F ) be a continuous `-adic representation such that τ (I)
is finite and ΦK acts semisimply. Then we can write, possibly after a finite extension F 0
of the field F , τ in the form
M
τ=
ρi ⊗ χi
i
where the ρi are `-adic representations GK → GLd (F 0 ) which factors through a finite
quotient and the χi are unramified 1-dimensional `-adic representations.
Proposition 2. Let k be a field of characteristic p ≥ 0 and V a finite dimensional vector
space over k. Let τ : G → GL(V ) be a representation of G for which there exists a finite
index subgroup H of G, with p - [G : H], such that ρ|H is semisimple. Then ρ is semisimple.
Proof. Let W be a subrepresentation of G. Our aim is to find a complementary subrepresentation. Since ρ|H is semisimple, we can find a k[H]-submodule U such that
V = W ⊕ U , as k[H]-submodules. This means that we have k[H]-homomorphism of
modules f : V /W → V which is a section of π : V → V /W . We construct a k[G]-module
homomorphism g : V /W → V by using Maschke’s trick and show that it is a section of
π.
Proof of theorem. Assume that τ is absolutely irreducible.
Let L be a finite totally ramified extension of K such that τ (Gal(L̄/Lnr )) = 1. Say it
is of degree e, which is also the degree of the extension Lnr /K nr . Denote by L0 the Galois
closure of L.
The Galois group Gal(Lnr /L0 ) is topologically generated by the Frobenius element
ΦL0 . We aim to prove that ΦL0 lies in the centre of G := Gal(Lnr /K). Writing H
for Gal(Lnr /K nr ), we observe that G is topologically generated by ΦL and H. Also, if
f := [L0 : K], then ΦL0 = ΦfK . Thus, ΦL0 commutes with ΦL . On the other hand, notice
that both H and Gal(Lnr /L0 ) are normal in G. Therefore, the commutator [σ, ΦL0 ], where
σ is an element of H, lies in both H and Gal(Lnr /L0 ). But since L0 K nr = Lnr , we conclude
that this commutator must be 1, which means that ΦL0 commutes with H. It is therefore
central in G.
Using Schur’s lemma we now conclude that τ (ΦL0 ) = λId, for some λ in a finite
extension
√
f
of F . Now define the character χ by requiring χ(IK ) = 1 and χ(ΦK ) = λ. Set ρ :=
τ ⊗ χ−1 . It is now immediate to verify that this ρ is precisely the Galois representation we
were looking for.
Theorem 3. Let τ : GK → GLd (F ) be a continuous `-adic representation for which every
0
choice of Frobenius element acts semisimply on τ I , for any finite index subgroup I 0 of the
inertia group I. Then τ is of the form
1
2
τ=
M
ρi ⊗ sp(ni )
i
where the ni are integers and ρi is a continuous `-adic representation with finite image
of inertia and with Frobenius acting semisimply.
Note that continuity implies that there is a finite Galois
L of K such that
S∞ extension
√
n
τ (IL ) = τ (H), where H := Gal(L` /Lnr ) ∼
= Z` and L` := n=1 Lnr ( ` πL ). It is also easy
to check that we have the relation σΦK = ΦK σ q , q := #FK , for any σ ∈ H.
Lemma 4. For any σ ∈ Gal(L` /Lnr ) and θ ∈ Gal(L` /K nr ), we have σθ = θσ.
Proof. Indeed, we just need to check that the image of σ commutes with the image of
Gal(K̄/K tr ), since Gal(K tr /K nr ) is pro-cyclic. But notice that while Gal(K̄/K tr ) is a
p-group, Gal(K̄/Lnr ) is an `-group. This means that the intersection of these two subgroups is trivial. However, picking an element σ ∈ Gal(K̄/Lnr ) and another element
θ ∈ Gal(K̄/K tr ), we have [σ, θ] lying in the intersection of these two groups. Therefore,
σθ = θσ.
Proof of case d = 1. Let us first prove the case d = 1. Our aim is to prove that we must
have finite inertia image. Pick some σ ∈ H and some Frobenius element ΦL over L. We
have the relation σΦL = ΦL σ q , where q := #FL . But then
τ (σ)q = τ (σ q ) = τ (Φ−1
L σΦL ) = τ (σ)
and so τ (σ) is a (q − 1)th root of 1. Thus, the image of inertia must be finite.
Proof of case d = 2. Let σ ∈ H be a (topological) generator of H. By extending the field
F if needed, we can put τ (σ) in its Jordan normal form. We now have three distinct
situations: either
λ 0
τ (σ) =
, λ 6= 0
0 λ
or
λ 0
τ (σ) =
, λ, µ 6= 0 and λ 6= µ
0 µ
or
λ 1
τ (σ) =
, λ 6= 0
0 λ
The first case reduces to the case d = 1 above and we conclude that the image of inertia
must be finite.
Consider now the second case. Let v1 = (1, 0) and v2 = (0, 1) and set V1 := hv1 i,
V2 := hv2 i. Note that the lemma above implies that τ (θ) is diagonal for any θ ∈ IK . Recall
that we have the relation σΦK = ΦK σ q , for any σ ∈ H, for any choice ΦK of Frobenius
element over K and where q = #FK . This implies that ΦK Vi is an H-subrepresentation.
Therefore, either we have ΦK V1 = V1 , ΦK V2 = V2 for every choice of Frobenius element
or we have ΦK V1 = V2 , ΦK V2 = V1 for every choice of Frobenius (as inertia does not
swap these two subspaces). If ΦK V1 = V2 , then, for any v1 ∈ V1 − {0}, we have µ(gv1 ) =
3
σ(gv1 ) = g(λr v1 ) = λr gv1 , where r = #FK . Therefore, λr = µ. On the other hand,
λgv2 = σgv2 = g(µr v2 ) = µr gv2 , for any v2 ∈ V2 − {0}, and so µr = λ. What this means
2
is that µr −1 = 1, and therefore both µ and λ are roots of unity and the image of inertia
is therefore finite. If ΦK Vi = Vi , i = 1, 2, then we conclude that τ is the direct sum of two
1-dimensional subrepresentations.
λ 1
The remaining case is when τ (σ) =
. As above, we conclude that for any
0 λ
ΦK ∈ GK the subspace ΦK V1 is a subrepresentation of τ |H. But in this case, since there
is only one 1-dimensional subrepresentation of τ |H, we see that we must have ΦK V1 = V1 .
Therefore, V1 is a subrepresentation of τ .
Assume now we are in this last case. Let χ be the 1-dimensional subrepresentation V1 .
Then, by the case d = 1, this has finite inertia image. Set τ 0 = τ ⊗ χ−1 . For any g ∈ GK ,
1 ∗
the image of g by τ 0 is of the form
. Note that, possibly after a finite field extension,
0 ∗
1 ∗
we may assume that τ 0 (σ) =
, for any σ ∈ Gal(L` /Lnr ).
0 1
1 ∗
0
This implies that τ (θ) = χ2 (θ)
, for any θ in the inertia group. Set τ 00 = τ 0 ⊗χ−1
2 .
0 1
1 1
We can now find a basis such that τ 00 (σ) =
, for any topological generator σ of
0 1
Gal(K` /K nr ).
Using now the fact that Φ−1 σΦ = σ q , where q = #FK and Φ is any choice of Frobenius
over K, we get what we wanted.