We now move on to our final task, showing that πF,pmust be induced from the
Borel subgroup of GSp4(Qp). In this section Λ0 will be an arbitrary prime of
K = QFQf, lying above a rational prime l0.
Recall that πF,pcorresponds via Local Langlands to a representation of the
Weil-Deligne group Wp0, which itself is parametrized by a continuous represen- tation ρ0 : WQp → GSp4(C) and a nilpotent matrix N ∈ M4(C) with certain
properties (mentioned in the previous subsection). However if we fix a choice of embeddings Q ,→ C and Q ,→ Ql0 then one can convert these representations
into l0-adic representations with open kernel.
It is also known that local Galois representations give rise to Weil-Deligne representations.
Theorem 5.2.5. (Grothendieck-Deligne) Let p 6= l0 and fix a continuous n- dimensional Λ0-adic representation:
ρ : Gal(Qp/Qp) −→ GLn(KΛ0).
Then associated to ρ is a unique l0-adic representation of W0
Qp, given by a
pair (ρ00, N0) satisfying:
• ρ0
0: WQp −→ GLn(KΛ0) is continuous with respect to the discrete topology
on GLn(KΛ0). In particular ρ00(Ip) is finite.
• ρ0
0(φp) has characteristic polynomial defined over GLn(OΛ0) with constant
term a unit. • N0∈ M
n(KΛ0) is nilpotent and satisfies
ρ00(σ)N0ρ00(σ)−1= ν(σ)N0, for all σ ∈ WQp.
5.2. Why paramodular?
Fixing the tamely ramified character tl0 : Ip → Zl0, the relationship between ρ
and ρ00 is:
ρ(φnpu) = ρ00(φnpu)exp(tl0(u)N0),
for all n ∈ Z, u ∈ Ip.
Now consider the local Galois representation ρF,p. By the above theorem
it has an associated Weil-Deligne representation, given by a pair (ρ00, N0). A Local-Global Compatibility conjecture of Sorensen (pages 3-4 of [61], proved in certain cases by Mok in Theorem 4.14 of [52]) predicts that the Weil-Deligne representations attached to πF,pand ρF,pare isomorphic (up to Frobenius semi-
simplification). In particular this implies that ρ0 ∼= ρ00 up to Frobenius semi-
simplification. We make this identification from now on and use ρ0 to denote
both representations.
A useful corollary of the above theorem is the following:
Corollary 5.2.6. (Grothendieck Monodromy Theorem) With the above setup there exists a finite index subgroup JΛ0 ⊆ Ip such that ρ(σ) = exp(tl0(σ)N ) for
each σ ∈ JΛ0, i.e. each element of JΛ0 acts unipotently.
See the appendix of [58] for a proof of this.
By the Grothendieck Monodromy Theorem there exists a (maximal) finite index subgroup JΛ0 ⊆ Ip acting by unipotent matrices, i.e. if σ ∈ JΛ0 then:
ρF,p(σ) = exp(tl0(σ)N ).
Note then that as a consequence, for each σ ∈ JΛ0:
ρ0(σ) = ρF,p(σ)exp(−tl0(σ)N ) = I.
Thus ρ0 factors through Ip/JΛ0:
ρ0: Ip−→ Ip/JΛ0 −→ GL4(OΛ0).
Note that ρ0(Ip/JΛ0) is finite. It is conjectured that the size of this image
is independent of Λ0 (by Conjecture 1.3 on p.81 − 82 of [63] one expects a compatible system of Galois representations).
A generalization of an argument on p.46−48 of [11] tells us about the possible sizes of this image.
Lemma 5.2.7. Suppose G is a finite subgroup of GLn(OΛ0) and that l0 > e + 1
(where e is the ramification index of KΛ0/Ql0). Then |G| divides |GLn(FΛ0)|.
Proof. We claim that the kernel of the reduction map GLn(OΛ0) → GLn(FΛ0)
Then restricting to G we must have trivial kernel (since G is finite), hence G injects into GLn(FΛ0).
To prove the claim we take A ∈ GLn(OΛ0) with A 6= I and A ≡ I mod Λ0.
We wish to prove that Am6= I for each m. We already know this for m = 1.
Suppose that A has finite order m > 1. Then choosing a prime q | m we see that (Ak)q = I (where m = qk). Letting B = Ak we see that B 6= I (since A has order bigger than k) and that B ≡ I mod Λ0. We have found a matrix
with the same conditions as A with prime order. Thus it suffices to show that no such matrix can have prime order.
To this end we write A = I + M with M 6= 0 and M having entries in Λ0 (since A ≡ I mod Λ0). Choose an entry mu,v of M such that |mu,v|Λ0 = δ is
maximal among all entries of M . Then 0 < δ ≤ N (Λ10)(normalizing the absolute
value in the usual fashion). Note that: Aq = (I + M )q = I + qM +q 2 M2+ ... + q q − 1 Mq−1+ Mq.
Case 1: Suppose q 6= l0. Then the entries of qjMj for j ≥ 2 all have Λ0-adic
absolute value less than or equal to δ2. However qM contains the entry qm u,v
of absolute value δ > δ2 (since q 6= l0). Hence Aq− I must contain an entry of
absolute value δ > 0 and so Aq− I 6= 0 as required.
Case 2: q = l0. We need sharper inequalities for this case since qM has no entry of absolute value δ. However it does contain the entry qmu,v of absolute
value δ
N (Λ0)e (since by definition the ramification index of the extension is e).
For 2 ≤ j ≤ q − 1 we know that q divides qj so the matrices q jM
j have
entries of maximal absolute value N (Λδ20)e ≤
δ N (Λ0)e+1 <
δ N (Λ0)e.
(We did not need to take into account divisibility of binomial coefficients in Case 1, weaker inequalities were enough.)
The matrix Mq has entries of absolute value greater than or equal to δq <
δe+1≤ δ
N (Λ0)e (using here the condition q = l0> e + 1).
Thus we see that Aq− I contains an entry of absolute value δ
N (Λ0)e > 0 and
so Aq− I 6= 0 as required.
Set N (Λ0) = l0f. Then a simple linear algebra argument shows: |GL4(FΛ0)| = (l04f− l03f)(l04f− l02f)(l04f − l0f)(l04f − 1)
5.2. Why paramodular?
Now by using what we know about ρF,pfrom the congruence we may prove
the following.
Theorem 5.2.8. If JΛ0 6= Ip and mΛ0 = |ρ0(Ip/JΛ0)| then l|mΛ0.
Proof. Note that mΛ0 = |ρF,p(Ip/JΛ0)|.
As mentioned it is conjectured that mΛ0 has order independent of Λ0. Thus
we may make the choice Λ0 = Λ.
Now G = ρF,p(Ip/JΛ) is a finite subgroup of GL4(OΛ). It must embed into
GL4(FΛ) by reduction (since by the proof of the above lemma G cannot contain
any non-trivial elements in the kernel of reduction). Thus |G| = |ρF,p(Ip/JΛ0)|.
However by the congruence we already know that the mod Λ reduction ρF,p has composition factors ρf,p, χk−2l , χj+k−1l .
Then since ρf,p=
1 ?
0 1
and χl is unramified at p we have:
ρF,p(Ip/JΛ) ⊆ 1 ? ? ? 0 1 ? ? 0 0 1 ? 0 0 0 1 .
However JΛ6= Ip, so that ρF,p(Ip/JΛ) is non-trivial, showing that N (Λ) divides
|G|. Thus l divides |G|. Then by the independence of Λ0mentioned above l|m Λ0
for any Λ0.
Corollary 5.2.9. Let KΛ/Ql have ramification index e and residue degree f .
If l ≥ max{6f + 2, e + 2} then JΛ0 = Ip for some Λ0.
Proof. Suppose JΛ0 6= Ip for all Λ0. Then we know that l|mΛ0 for all Λ0. But
for each Λ0 we know that ρ0(Ip/JΛ0) is a finite subgroup of GL4(OΛ0) and so
(restricting to those extensions such that l0> e + 1): l|l06f(l0f − 1)(l02f− 1)(l03f− 1)(l04f − 1), for all such l0.
It remains to prove that l0 can be chosen to contradict this. To contradict the divisibility condition it suffices to choose l06= l such that l03f 6≡ 1 mod l and
l04f 6≡ 1 mod l. To do this we will show that, under the condition on l, there exists a non-zero class mod l that does not have order dividing 3f or 4f .
Note that since l is prime there are at most n solutions to the congruence xn ≡ 1 mod l, hence at most n classes mod l of order dividing n. Thus there
note that the classes of order dividing hcf(3f, 4f ) = f are counted twice and so there must be at most 7f − f = 6f classes of order dividing 3f and 4f .
But since l ≥ 6f +2 there must be a non-zero class mod l that doesn’t satisfy these congruences. By Dirichlet’s theorem there are infinitely many primes in this class mod l.
It suffices to choose l0 to be in this class with l0 ≥ e + 2 (so that l0 > e + 1
too).
Of course it is highly likely that l ≥ max{6f + 2, e + 2} in practice since l is a “large” prime. The result will still hold true for certain l ≤ max{6f + 2, e + 2} but it is not so easy to find a good choice for l0.
It remains to prove that the case JΛ0 = Ip(known as “semi-stable” in Wiese
[69]) implies that πF,pis induced from the Borel subgroup of GSp4(Qp).
Proposition 5.2.10. If Λ0 satisfies JΛ0 = Ip then πF,p is induced from the
Borel subgroup of GSp(Qp).
Proof. It suffices to show that there is a basis of KΛ40 such that
ρ0∼= χ1 ? ? ? 0 χ2 ? ? 0 0 χ3 ? 0 0 0 χ4 ,
for four unramified characters χ1, χ2, χ3, χ4 of WQp. Then since the image of
ρ0 lies in GSp4 we must have that χ3 = χ−11 and χ4 = χ−12 . Then by Local
Langlands for GSp4 it must be that πF,pis induced from the Borel subgroup.
To this end we already know that Ip acts unipotently and so it remains
to study the action of Frobenius φp. Recall the condition ρ0(φp)N ρ0(φp)−1 =
p−1N . We will rewrite this as ρ
0(φp)N = p−1N ρ0(φp).
By Theorem 5.2.5 the characteristic polynomial of ρ0(φp) has constant term
in OΛ×0 we may choose an eigenvector v of ρ0(φp) with non-zero eigenvalue α ∈
OΛ0. Then notice that
ρ0(φp)(N v) = p−1N ρ0(φp) = αp−1(N v).
This shows that if N v 6= 0 then N v is another eigenvector of ρ0(φp) with
eigenvalue αp−16= α.
Consider the list v, N v, N2v, N3v. If all of these vectors are non-zero then
we have a basis of eigenvectors for ρ0(φp). Then ρ0(φp) is diagonal.
If for some i ≤ 3 we have Niv = 0 then we can quotient out by the subspace
generated by v, N v, ..., Ni−1v and choose another eigenvector w for ρ
0(φp) acting