Notes on Ribet’s Lemma
Mark Reeder
December 8, 2014
Let K be a locally compact field, complete with respect to a discrete valuation, with ring of integers
A, a fixed prime element π ∈ A and residue field k = A/πA. Let q = |k|.
Let V be a two-dimensional vector space over K. A lattice in V is a rank-two A-submodule
L⊂V.
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Reduction of representations
Suppose G is a compact topological group and we have a continuous representation
ρ : G → GL(V ).
We say that a lattice L ⊂ V is G-stable if ρ(G)L = L.
Proposition 1.1. There exists a G-stable lattice L ⊂ V .
Proof. Let L0 be any lattice in V , and write the elements of ρ(G) as matrices with respect to a
basis {u, v} of L0 . Since G is compact, these matrices are bounded. Hence there is an integer
n ≤ 0 such that ρ(g)L0 ⊂ π n L0 for all g ∈ G. The A-span of ρ(G)L0 lies between L0 and π n L0 ,
hence is a lattice.
From any G-stable lattice L in V , we get a representation
ρ̄L : G → GL(L/πL) ' GL2 (k),
whence an action of G on the projective space P(L/πL) ' P1 (k). Let f be the number of fixedpoints of G in P(L/πL). There are four possibilities:
1. ρ̄L is irreducible. (f = 0)
2. ρ̄L is reducible but indecomposable, meaning that it fits into a non-split extension
0 −→ χ −→ ρ̄L −→ η −→ 0.
(f = 1)
3. ρ̄L = χ ⊕ η is a direct sum of distinct characters χ, η : G → k × . (f = 2)
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4. ρ̄L = χ ⊕ χ is a direct sum of two copies of a single character χ : G → k × . (f = q + 1)
We have case 2 if and only if ρ̄L is reducible and p divides the order of ρ̄L (G). Indeed, in the
completely reducible cases 3,4 we have ρ̄L (G) ⊂ (k × )2 and conversely if p does not divide | im ρ̄L |
then ρ̄L is completely reducible, by Mackey’s theorem.
The structure of ρ̄L depends on L, as we will see. However its semisimplification does not depend
on L. That is, let ρ̄ss
L be the direct sum of the irreducible constituents of ρ̄L . A theorem of Brauer
and Nesbitt (reference [4,Th. 30.16] in Ribet) says that ρ̄ss
L depends only on the function g →
ss
det(X − ρ̄L (g)). As this polynomial is the reduction mod π of det(X − ρ(g)), it follows that ρ̄ss
L
is independent of L. We henceforth denote this representation by ρ̄ss .
“Ribet’s Lemma" is the following.
Proposition 1.2 (Ribet, Prop. 2.1). If ρ is irreducible but ρ̄ss is reducible, then there exists a
G-stable lattice L ⊂ V such that ρ̄L is indecomposable (case 3 above).
We will give Serre’s proof of this result, which is based on properties of the Bruhat-Tits tree.
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The Bruhat-Tits tree
If L is a lattice in V and a ∈ K × , then aL is another lattice. Thus, the group K × acts on the
set of lattices in V . Let X be the quotient space {lattices in V }/K × , and let [L] ∈ X denote the
K × -orbit of a lattice L ⊂ V .
We define a graph X with vertex set X as follows. Two points x = [L] and x0 = [L0 ] in X are
connected by an edge if πL ⊂ L0 ⊂ L. One checks this is well-defined and symmetric in x, x0 .
Given two lattices L, L0 in V , the theory of elementary divisors shows there is a basis {u, v} of L
and integers i, j such that {π i u, π j v} is a basis of L0 . One checks that the distance function
d(x, x0 ) = |i − j|
is well-defined. Let x = [L]. Then x0 has a unique representative lattice L0 such that L0 ⊂ L but
L0 6⊂ πL. One checks that L/L0 ' A/π n A, where n = d(x, x0 ). In particular we have d(x, x0 ) = 0
if and only if x = x0 and d(x, x0 ) = 1 if and only if {x, x0 } is an edge in X.
Proposition 2.1. The graph X is connected and simply-connected. In other words, between any
two vertices in X there is a unique path in X with no repeated vertices.
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Proof. (from Serre’s Trees) Given x, x0 ∈ X, choose lattices L ∈ x, L0 ∈ x0 such that L0 ⊂ L.
Consider a composition series of A-modules
L0 = Ln ⊂ Ln−1 ⊂ · · · ⊂ L0 = L.
For each 1 ≤ i ≤ n we have Li−1 /Li ' k, so that πLi−1 ⊂ Li ⊂ Li−1 . Setting xi = [Li ], we
obtain a path from x = x0 to x0 = xn . Removing sub-paths between repeated vertices if necessary,
we obtain a path from x to x0 without repeated vertices.
For simply-connectedness, it suffices to show that if x = x0 , x1 , . . . , xn = x0 is a path without
repeated vertices then x 6= x0 . In fact, we show that d(x, x0 ) = n > 0, using induction on n.
Choose Li ∈ xi such that Li ⊂ Li−1 but Li 6⊂ πLi−1 . Then L0 /Ln ' A/π n A. We will have
d(x, x0 ) = n if and only if Ln 6⊂ πL0 . Since xn , xn−1 , xn−2 is a path with three distinct points, we
have
πLn−1 ⊂ Ln ⊂ Ln−1 ,
and
πLn−1 ⊂ πLn−2 ⊂ Ln−1
and Ln and πLn−2 project to distinct lines in Ln−1 /πLn−1 . Hence we have Ln−1 = Ln + πLn−2 .
By induction we have Ln−1 6⊂ πL0 , so Ln 6⊂ πL0 , as desired.
Thus, X is a tree (connected graph without cycles) such that the neighbors of each vertex x = [L]
correspond to points in the projective space P(L/πL).
The group GL(V ) acts transitively on lattices in V , hence on X, and preserves the edge relations.
Thus GL(V ) acts on the tree X. The stabilizer of a vertex x = [L] is GL(L), which acts transitively
on the edges meeting x, since GL(L/πL) is transitive on P(L/πL). Hence X is a homogeneous
tree for GL(V ), where the degree of each vertex is |k| + 1.
Here is a picture of X, when |k| = 2. The shading of vertices indicates that SL(V ) has two orbits
on X.
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The tree X is infinite, and is in particular noncompact. It can be compactified with boundary P(V ),
as follows. A half-line in X is an infinite sequence h = (x0 , x1 , x2 , x3 , . . . ) of distinct vertices. We
call x0 the origin of h. As before, we may choose representatives Ln ∈ xn such that π n L0 ⊂ Ln
and L0 /Ln ' A/π n A. Set
\
`0 =
Ln .
n
Then
`0 = lim A/π n A ' A
←−
n
is a nonzero rank-one submodule of L0 and ` := K`0 is a line in V . We call ` the end of h.
Conversely, given a line ` ∈ P(V ) and a point x = [L] ∈ X, let `0 = L ∩ `. This is a rank-one
A-submodule of ` and we set xi = `0 + π n L.T hisgivesahalf − linewithoriginx= Thus P(V )
may be regarded as the boundary of a compactification of X.
This is analogous to P1 (R) being the boundary of the symmetric space X(R) = P1 (C)\P1 (R).
Likewise, the stabilizer in GL(R) of a point x ∈ X(R) acts transitively on spheres centered at x,
just as for x = [L] in the p-adic case the stabilizer GL(L) acts transitively on each metric sphere
{x0 ∈ X : d(x, x0 ) = n} = P1 (A/π n A).
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Proof of Ribet’s lemma
Return to our representation ρ : G → GL(V ) with G compact. Ribet’s lemma will become
transparent once we study the fixed-point graph XG consisting of the fixed-vertices X G = {x ∈
X : ρ(g)x = x ∀ g ∈ G} and all the edges of X between vertices of X G .
We have seen in Prop. 1.1 that X G is nonempty. Note also that if x = [L], then x ∈ X G if and
only if L is G-stable. (One direction is clear; for the other direction, note that if ρ(g)L = π i L, then
i = 0, since ρ(G) is bounded.)
Proposition 3.1. The graph XG is connected.
Proof. Let x, x0 ∈ X G , and let x = x0 , x1 , . . . , xn = x0 be the unique path in X from x to x0 .
We must show that every xi ∈ X G . But if g ∈ G, applying ρ(g) to this path gives a new path
from x to x0 , since x, x0 are fixed. Since the path is unique, we must have ρ(g)xi = xi for each
0 ≤ i ≤ n.
We next show that many properties of ρ, and its reductions, are reflected in the structure of XG .
Proposition 3.2. The graph XG is bounded if and only if ρ is irreducible.
Proof. If XG is not bounded then it contains a half-line h = (x0 , x1 , . . . ) whose end ` ∈ P(V ) is
also fixed by ρ(G). Hence ρ is reducible. Conversely, suppose ` is a G-stable line in V . Choose
any vertex x0 = [L] ∈ X G and form the half-line h = (x0 , x1 , . . . ) in X. Each xn = (L ∩ `) + π n L
is G-stable, so h is a half-line in XG . Hence XG is unbounded.
We turn now to the reduction ρ̄L : G → GL(L/πL) at a fixed-point x = [L] ∈ XG . The degree of
x in XG is the number f of fixed-points of G in P(L/πL). From the classification in section 1, the
cases are as follows.
1. ρ̄L is irreducible if and only if XG = {x} is a single point.
2. ρ̄L is reducible and indecomposable if and only if x has exactly one neighbor in XG . (We
say x is a leaf.)
3. ρ̄L is a direct sum of two distinct characters if and only if x has exactly two neighbors in XG .
4. ρ̄L is a scalar representation if and only if three neighbors of x belong to XG (in which case
all q + 1 neighbors are in XG ).
Now to prove Ribet’s lemma. Suppose ρ is irreducible, but some (any) reduction ρ̄L is reducible.
The fixed-point set XG is a tree. Recall the well-known theorem: Trees have leaves. Let x ∈ XG
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be a leaf. Then ρ̄L is indecomposable, by case 2 above. Thus Ribet’s theorem is proved.
But we want more: if ρ̄L is indecomposable there could be two possible extensions, and we want
to show that both extensions occur for suitable choices of L. This only matters if the composition
factors of L/πL are non-isomorphic. This condition excludes case 4 from occuring at any vertex
in XG , so each vertex in XG has at most two neighbors. Since XG is connected, this means it is a
segment with exactly two leaves x, x0 :
x = x0 —x1 — · · · —xn−1 —xn = x0 ,
where x0 , . . . , xn are the distinct vertices in XG . We have n ≥ 1 since ρ has reducible reductions.
Let xi = [Li ], where πLi−1 ⊂ Li ⊂ Li−1 for 1 ≤ i ≤ n.
Lemma 3.3. As G-representations we have
L0 /L1 ' Li−1 /Li ,
for all 1 ≤ i ≤ n.
Proof. Note that L0 /L1 is the quotient in the exact sequence
0 −→ L1 /πL0 −→ L0 /πL0 −→ L0 /L1 −→ 0.
We may assume n ≥ 2 and that the isomorphism in the lemma holds for some 1 ≤ i < n. We must
show that it holds for i + 1.
The lattice Li contains both Li+1 and πLi−1 , and these both contain πLi . Since xi+1 6= xi−1 , we
have Li+1 6= πLi−1 , so they project to distinct G-stable lines in Li /πLi . Thus, we have
Li /πLi ' (Li+1 /πLi ) ⊕ (πLi−1 /πLi ).
now
πLi−1 /πLi ' Li−1 /Li ' L0 /L1 ,
by the induction hypothesis. This is the quotient of L0 /πL0 . Hence Li+1 /πLi is the submodule of
L0 /πL0 :
Li+1 /πLi ' L1 /πL0 .
But Li+1 /πLi is also the quotient in the exact sequence
0 −→ πLi /πLi+1 −→ Li+1 /πLi+1 −→ Li+1 /πLi −→ 0.
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It follows that
πLi /πLi+1 ' L0 /L1 .
But πLi /πLi+1 ' Li /Li+1 , so we have shown that Li /Li+1 ' L0 /L1 , as desired.
At the other end of the segment we have the exact sequence
0 −→ Ln−1 /Ln −→ Ln /πLn −→ Ln /πLn−1 .
From the lemma we immediately get the following.
Proposition 3.4. The unique G-quotient of L0 /πL0 is isomorphic to the unique G-submodule of
Ln /πLn .
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Ribet’s use of his lemma...
... is in Prop. 4.2. By this point in the paper, he has a p-adic field K (where p | Bk−1 ), a twodimensional K-vector space V and an irreducible representation ρ : GQ → GL(V ) which for
every prime ` 6= p satisfies
tr(ρ(Frob` ) ≡ 1 + `k−1
mod π
and
det(Frob` ) ≡ `k−1
mod π.
Note this makes sense because GQ , being compact, stabilizes a lattice L ⊂ V , so tr(ρ(g)) and
det(ρ(g)) lie in the ring of integers A ⊂ K, for all g ∈ GQ .
Now the cyclotomic character χ satisfies χ(Frob` ) ≡ ` mod π. It follows that the identity
det(X − ρ(g)) ≡ (X − 1)(X − χk−1 (g))
holds for all Frob` , ` 6= p. The union of the conjugacy-classes of these Frobenius elements is dense
in GQ , by the Chebotarev Theorem for infinite extensions (sec. 1.2.2 of reference [17] in Ribet). It
follows that the identity holds for all g ∈ GQ .
From the Brauer-Nesbitt theorem quoted earler, it follows that ρss ' 1 ⊕ χk−1 . By Ribet’s lemma
there exists a lattice L ⊂ V which is indecomposable, and by Prop. 3.4 we can choose L such that
ρ̄L fits into the exact sequence
0 −→ 1 −→ ρ̄L −→ χk−1 −→ 0,
as asserted in Ribet’s Prop. 4.2.
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