On Ribet's Lemma for $\mathrm{GL}_2$ modulo prime powers

Let $\rho\colon G\to \mathrm{GL}_2(K)$ be a continuous representation of a compact group $G$ over a complete discretely valued field $K$, with ring of integers $\mathcal O$ and uniformiser $\pi$. We prove that $\operatorname{tr}\rho$ is reducible modulo $\pi^n$ if and only if $\rho$ is reducible modulo $\pi^n$. More precisely, there exist characters $\chi_1,\chi_2 \colon G\to(\mathcal O/\pi^n\mathcal O)^{\times}$ such that $\det(t - \rho(g))\equiv (t-\chi_1(g))(t-\chi_2(g))\pmod{\pi^n}$ for all $g\in G$, if and only if there exists a $G$-stable lattice $\Lambda\subset K^2$ such that $\Lambda/\pi^n\Lambda$ contains a $G$-invariant, free, rank one $\mathcal O/\pi^n\mathcal O$-submodule. Our result applies in the case that $\rho$ is not residually multiplicity free, in which case it answers a question of Bella\"iche--Chenevier. As an application, we prove an optimal version of Ribet's Lemma, which gives a condition for the existence of a $G$-stable lattice $\Lambda$ that realises a non-split extension of $\chi_2$ by $\chi_1$


Introduction
Let O be a complete discrete valuation ring with fraction field K, uniformiser π, discrete valuation v π normalised such that v π pπq " 1 and residue field F. Let G be a compact group and let ρ : G Ñ GL 2 pKq be a continuous representation.Suppose that there exists an integer n and continuous characters χ 1 , χ 2 : G Ñ pO{π n Oq ˆsuch that, for all g P G, we have (1.1)P ρpgq ptq :" detpt ´ρpgqq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q, where P ρpgq ptq is the characteristic polynomial of ρpgq.The goal of this paper is to answer the following questions, the first of which is equivalent to [BC14b, Question, pp.524]: Question 1.Is ρ reducible modulo π n , i.e. does there exist a G-stable lattice Λ Ď K 2 such that Λ{π n Λ contains a G-invariant, free, rank one O{π n O-submodule V ?
Equivalently, does there exist a basis for K 2 with respect to which the image of ρ is a subgroup of Γ 0 pπ n q :" #˜a b c d ¸P GL 2 pOq : c " 0 pmod π n q + ?Question 2. Can we choose Λ so that the rank one submodule V is isomorphic to χ 1 ?Moreover, if ρ is irreducible, can we choose Λ so that Λ{π n Λ is a non-split extension of χ 2 by χ 1 ?
Question 1 has previously been studied by Katz [Kat81] in the context of Galois representations attached to elliptic curves, and answered in the case that K is a finite extension of Q p and χ 1 is the trivial character [Kat81, Thm.1], though his proof can be generalised to any χ 1 , χ 2 .Our proof is completely different and works in the more general case of discretely valued fields.Moreover, our argument only requires G to be a semigroup.
1.1.The Bruhat-Tits tree of PGL 2 pKq.In order to answer these questions, we rephrase them in the language of Bruhat-Tits trees.
By a lattice in K 2 we mean a rank two O-module Λ Ď K 2 that spans K 2 as a vector space.We say that two lattices Λ 1 , Λ 2 are homothetic if there is an element a P K such that Λ 1 " aΛ 2 .
Definition 1.1.The Bruhat-Tits tree X of PGL 2 pKq is the graph whose vertices are the homothety classes of lattices in K 2 .Two vertices x, y P X are joined by an edge if we can choose representatives Λ x , Λ y of x, y such that Equivalently, x, y P X are neighbours if there exists a basis pv 1 , v 2 q of Λ x such that the homothety class of the lattice with basis pv 1 , πv 2 q is y.After fixing such a basis, the other neighbours of Λ x are the lattices with bases pπv 1 , v 2 `iv 1 q where i P O runs over a set of representatives for the congruence classes of O modulo π.Thus, if q denotes the (possibly infinite) cardinality of F, then X is a pq `1q-regular tree.
By extension, suppose that x, y P X are two vertices of X of distance d " dpx, yq from each other, where dpx, yq is the number of edges in the path connecting x to y.Then d is the smallest integer for which we can choose representatives Λ x , Λ y such that Equivalently, d is the unique integer for which there exists a basis pv 1 , v 2 q of Λ x such that pv 1 , π d v 2 q is a basis of Λ y .
The representation ρ induces an action of G on the vertices of X .Let X pρq be the subgraph of X whose vertices are homothety classes of G-stable lattices.Then, for each x P X pρq with representative Λ x , there is a bijection between vertices y P X pρq with dpx, yq " n and free, rank one, G-invariant O{π n O-submodules of Λ x {π n Λ x (Lemma 2.3).Thus, an affirmative answer to Question 1 follows from the following theorem, which is our main result: Theorem 1.2.The following are equivalent: piq There exist two vertices x, y P X pρq with dpx, yq " n.
Questions 1 and 2 have been answered in far greater generality in the case that ρ is residually multiplicity-free, i.e. if χ 1 ı χ 2 pmod πq [Bel03, BG06, BC09, Che14].In the residually multiplicity-free case, the graph X pρq is just a finite line segment, and the theory is simplified considerably by the fact that the characters χ 1 , χ 2 , if they exist, are unique.
However, the non-residually multiplicity-free case is far more mysterious.This case has been studied by Bellaïche-Chenevier [BC14b], who classify the types of graphs X pρq can be, prove that piq implies piiq in Theorem 1.2 and give a partial result in the opposite direction [BC14b,Thm. 45].They pose as a question whether piq and piiq are equivalent [BC14b,Question,pp. 524].As well as answering this question, we complete Bellaïche-Chenevier's description of the shape of X pρq: we describe the shape of X pρq in terms of certain pseudocharacter invariants of ρ (Theorem 5.1).Conversely, we also show how to compute these invariants from the shape of X pρq (Theorem 5.3).
The non-residually multiplicity-free case is, again, more mysterious.Indeed, the naïve generalisation of Ribet's lemma is false: in Section 4.2, we give examples of representations ρ : G Ñ GL 2 pKq with P ρpgq ptq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q, but such that there is no G-stable lattice Λ Ď K 2 for which either of χ 1 , χ 2 is a submodule of Λ{π n Λ.Note that, if ρ is irreducible, then Theorem 1.2 shows that there do exist pairs of characters η 1 , η 2 and G-stable lattices Λ such that Λ{π n Λ is a non-split extension of η 2 by η 1 .The problem is that, unlike in the multiplicity-free case, the decomposition of P ρpgq pmod π n q as a product of characteristic polynomials of characters is not unique, and only some of the decompositions can actually be realised by lattices.
Our first application of Theorem 1.2 is the following generalisation of Ribet's Lemma, which is optimal, in the sense that the integer s is as large as possible in general: Theorem 1.3.Let ρ : G Ñ GL 2 pKq be an irreducible representation and let χ 1 , χ 2 : G Ñ pO{π n Oq ˆbe characters such that, for all g P G, the characteristic polynomial P ρpgq ptq of ρpgq factors as pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q.Let m " mpχ 1 , χ 2 q be the largest integer such that χ 1 " χ 2 pmod π m q.Define s " spχ 1 , χ 2 q " Then there exists a G-stable lattice Λ such that Λ{π s Λ is a residually non-split extension of χ 1 pmod π s q by χ 2 pmod π s q.
Here, by a residually non-split extension, we mean that Λ{πΛ is indecomposable.We note in Corollary 2.14 that the integer s depends only on ρ and n and not on the choice of characters χ 1 , χ 2 .1.3.Isogenies of elliptic curves.Let F be a number field and let E{F be an elliptic curve.If p is a prime of good reduction for E and if the absolute ramification index e p of p satisfies e p ă p ´1, then there is an injective map EpF q tors ãÑ EpF p q from the torsion subgroup of EpF q to the points of E over the residue field F p .In particular, if ℓ is a prime, n ě 1 and ℓ n | #EpF q tors , then ℓ n | #EpF p q for all primes p of good reduction.
In [Kat81], Katz studied Question 1 in order to prove a converse to this statement.Let ρ ℓ : GalpF {F q Ñ GL 2 pQ ℓ q be the ℓ-adic Galois representation attached to (the isogeny class of) E. Then lattices inside ρ ℓ are in bijective correspondence with elliptic curves E 1 that are ℓ-power isogenous to E, and the ℓ-isogeny graph of E is exactly the invariant subtree X pρ ℓ q.By the Chebotarev density theorem and the definition of ρ ℓ , the condition that ℓ n | #EpF p q for all primes p of good reduction is equivalent to ρ ℓ satisfying detp1 ´ρℓ pgqq " 0 pmod ℓ n q for all g P GalpF {F q, which is equivalent to p1.1q with χ 1 " 1 and χ 2 the mod ℓ n cyclotomic character.By answering Question 1 with K " Q p and χ 1 " 1, Katz showed that if ℓ n | #EpF p q for all primes p of good reduction, then there is an elliptic curve E 1 , isogenous to E over F , such that ℓ n | #E 1 pF q tors .Similarly, if E admits a cyclic ℓ n -isogeny over F , then there is a character χ : GalpF {F q Ñ pZ{ℓ n Zq ˆsuch that χpgq is a root of the characteristic polynomial P ρpgq ptq modulo ℓ n , for all g P GalpF {F q.Moreover, if E has good reduction at a prime p ∤ ℓ, then χ is unramified at p. On the other hand, E is isogenous over F to an elliptic curve E 1 that admits a cyclic ℓ n -isogeny if and only if there is a line of distance n in its ℓ-isogeny graph, which is exactly X pρ ℓ q.Hence, as an immediate consequence of Theorem 1.2, we deduce the following corollary: Corollary 1.4.Let E be an elliptic curve over a number field F , and let S be the set of primes of bad reduction for E and the primes above a rational prime ℓ.Let ρ ℓ : GalpF {F q Ñ GL 2 pQ ℓ q be the ℓ-adic Galois representation attached to E. The following are equivalent: piq E is isogenous over F to an elliptic curve E 1 that admits a cyclic ℓ n -isogeny.
piiq There exists a character χ : GalpF {F q Ñ pZ{ℓ n Zq ˆ, unramified outside S, such that χpFrob p q is a root of the characteristic polynomial P ρ ℓ pFrobpq ptq modulo ℓ n , for set of primes p of F of Dirichlet density 1.
Note that, if in part piiq, we instead just assume that the characteristic polynomial of ρ ℓ pFrob p q is reducible modulo ℓ n for each p, then the equivalence is false.Indeed, this latter condition is equivalent to E being isogenous to an elliptic curve E 1 that admits a cyclic ℓ n -isogeny everywhere locally.However, there exist elliptic curves with no global isogenies that have isogenies everywhere locally [Sut12,Ann14,BC14a,Vog20]. We note that, since det ρ ℓ is the ℓ-adic cyclotomic character, the assumption b `´1 ℓ ˘ℓ R F imposed in [Sut12] and [Ann14] forces ρ ℓ to be residually multiplicity-free.

Preliminaries
In this section, we recall key properties of the Bruhat-Tits tree X of PGL 2 pKq and, for a representation ρ : G Ñ GL 2 pKq, we define the ρpGq-invariant subtree X pρq and discuss its shape.We then discuss the decompositions of ρ modulo π n and define the index of irreducibility npρq and the index of irreducibility with multiplicity mpρq.Our key reference is [BC14b, §2] (see also [Ser80]).
2.1.The ρpGq-invariant subtree.Let ρ : G Ñ GL 2 pKq be a representation of a group G and assume that there exists a ρpGq-stable lattice Λ Ď K 2 .In particular, for every g P G the characteristic polynomial P ρpgq ptq of g is an element of Orts.This condition is automatically satisfied for most representations of interest, for example, if G is compact and ρ is continuous.
Choosing a basis for Λ, we obtain a representation ρ Λ : G Ñ GL 2 pOq Ď GL 2 pKq that is isomorphic to ρ.Hence, we can define the residual representation ρ Λ : G Ñ GL 2 pFq as well as the mod π n representations ρ Λ pmod π n q.In general, the isomorphism class of the representation ρ Λ depends on the homothety class of Λ.
Definition 2.1.Let A Ď M 2 pKq be a set of matrices.We denote by X pAq the induced subtree of X of all A-stable vertices, i.e. vertices x P X for which aΛ x Ď Λ x for all a P A and for some (and hence any) representative Λ x .We define X pρq to be X pρpGqq.

Remarks 2.2. piq If
piiq If a, b P M 2 pKq and λ, µ P O, then X pta, buq " X pta, b, abuq " X pta, b, λa `µbuq.Hence, if A Ď M 2 pKq is a finitely-generated O-algebra, then we can compute X pAq by computing X pta 1 , . . ., a n uq for an explicit set of generators.
The following lemma shows the relationship between the structure of the graph X pρq and subrepresentations of ρ Λ pmod π n q: Lemma 2.3 ([BC14b, Prop.11]).Let x P X pAq and fix a representative Λ x of x.There is a bijection between ‚ Points y P X pAq with dpx, yq " n; ‚ Free, rank one O{π n O-submodules of Λ x {π n Λ x that are A-stable.
Given a point y P X pAq with dpx, yq " n, the corresponding submodule of Λ x {π n Λ x is given by Λ y {π n Λ x , where Λ y is chosen so that Remark 2.4.Note that Lemma 2.3 immediately shows that Theorem 1.2 is equivalent to Question 1.Moreover, we see that: piq The tree X pρq is bounded if and only if ρ is an irreducible representation [BC14b, Lem.10].
piiq X pρq consists of a single point x if and only if ρ Λx is irreducible.
piiiq A vertex x P X pρq is a leaf, i.e. a vertex with exactly one neighbour, if and only if Λ x {πΛ x is indecomposable.
To describe this classification we recall their terminology.
Definition 2.5.Let S be a line segment in X , and let r be a positive integer.The band BpS, rq with nerve S and radius r is the subtree of X consisting of all vertices x with dpx, Sq ď r.
We distinguish two particular types of bands: ‚ If S " txu is a single vertex, Bpx, rq :" Bptxu, rq is the ball of radius r and centre x.
‚ If S " tx, yu consists of two adjacent vertices, we call BpS, rq a generalised ball.One can think of BpS, rq as a ball of radius r `1 2 around the middle of the segment rx, ys.
Definition 2.6.Suppose that ρ is irreducible, so that X pρq is a band.Define dpρq to be the diameter of X pρq and rpρq to be its radius.
By definition, the diameter of a tree is the length of a maximal path.The diameter of the band BpS, rq is therefore ℓpSq `2r, where ℓpSq is the length of the nerve S. To further illustrate the connection between the shape of X pρq and the representation theory of ρ, we record the following useful lemma: Lemma 2.7.Fix A Ď M 2 pKq and let H be the semigroup generated by A. Fix x P X pAq and r ě 1.Then H acts by a one-dimensional character on Λ x {π r Λ x if and only if Bpx, rq Ď X pAq.
Proof.Note that H acts as a one-dimensional character on Λ x {π r Λ x if and only if every free rank one O{π r O-submodule of Λ x {π r Λ x is stable under the action of H.By Lemma 2.3, this latter condition is equivalent to Bpx, rq Ď X pHq.By Remark 2.4, X pAq " X pHq.□ 2.3.Conjugate characters modulo π n .Let ρ : G Ñ GL 2 pKq be a representation.
Remark 2.9.When the residue characteristic of K is not 2, by the identity trpAq 2 ´trpA 2 q " 2 detpAq for all A P M 2 pKq, the above condition is equivalent to tr ρpgq " χ 1 pgq `χ2 pgq pmod π n q for all g P G.
If ρ is not residually multiplicity-free, then, in general, there exist multiple distinct pairs of conjugate characters modulo π n for every n ě 2. The following lemma shows that any two pairs of conjugate characters agree modulo π s , where s is as in Theorem 1.3.Lemma 2.12.Let pχ 1 , χ 2 q and pη 1 , η 2 q be two pairs of conjugate characters modulo π n , for some n.Let Then, up to reordering, we have χ 1 " η 1 pmod π s q and χ 2 " η 2 pmod π s q.
We think of npρq as the index of reducibility of ρ and of mpρq as the index of reducibility with multiplicity.
Remark 2.18.Define npρpGqq to be the largest integer n for which there exists a pair of conjugate characters pχ 1 , χ 2 q modulo π n , such that χ 1 , χ 2 factor through ρpGq.Then it is clear that npρpGqq ď npρq.However, in general there do exist pairs of conjugate characters that do not factor through the image of ρ, so a priori, it is not clear that these invariants should be equal.However, it follows from Theorem 1.2 that indeed npρpGqq " npρq " dpρq.Note that in [BC14b], the authors only consider npρpGqq.
Definition 2.19.Let U Ď X be a subset of the tree X .We say that a vertex x P U is an interior point of U if Bpx, 1q Ď U .Otherwise, we say that x is a boundary point of U .
Example.Suppose that U " BpS, rq is a band with nerve S and radius r.If r " 0, then U is a line segment and any vertex x P U is a boundary point.If r ą 0, the boundary points of U are exactly its leaves.
The following lemma is a slight generalisation of [BC14b, Lem.36]: Lemma 2.20.Let g P M 2 pOq and let x be a boundary point of X ptguq.Assume that the characteristic polynomial P g ptq of g factors as P g ptq " pt ´αqpt ´βq pmod π n q, for some α, β P O{π n O. Then there exists a basis pv 1 , v 2 q of Λ x {π n Λ x with respect to which g pmod π n q is represented by the matrix ˜α 1 Proof.Recall that, by Definition 2.1, X ptguq is the subtree of X consisting of homothety classes x of lattices Λ x such that gΛ x Ď Λ x .Note that, by Lemma 2.7, x is an interior point of X ptguq if and only if g acts by a scalar on Λ x {πΛ x Suppose that x is a boundary point.Since g does not act by a scalar on Λ x {πΛ x , there exists an element v P Λ x {πΛ x such that pv, gvq is a basis of Λ x {πΛ x .By Nakayama's lemma (for the ring O), if v P Λ x {π n Λ x lies above v, then pv, gvq is basis of Λ x {π n Λ x .Denote w 2 " v and let w 1 P Λ x {π n Λ x be the vector w 1 " pg ´βq ¨w2 .

□
Corollary 2.21.Let g P M 2 pOq, and let x be a boundary point of X ptguq.Let n be a positive integer.The following are equivalent: piq The characteristic polynomial P g ptq is reducible modulo π n .
piiq There exists a point y P X ptguq with dpx, yq " n.
Proof.Assume piq.Let pv 1 , v 2 q be a basis of Λ x {π n Λ x as in Lemma 2.20.Then v 1 spans a g-invariant rank one free submodule of Λ x {π n Λ x , which by Lemma 2.3 corresponds to a point y P X ptguq at distance n from x.
Assume piiq.Choose Λ y such that x is a rank one free submodule, stable under the action of g.Thus, there exists a basis pv 1 , v 2 q of Λ x {π n Λ x with respect to which g is upper triangular.Hence, P g ptq is reducible modulo π n .□

Proof of Theorem 1.2
Recall that ρ : G Ñ GL 2 pKq is a representation, with G a compact group.Let pχ 1 , χ 2 q be a pair of conjugate characters modulo π n .By definition, the characteristic polynomial P ρpgq ptq of ρpgq factors as P ρpgq ptq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q, for all g P G.
To prove Theorem 1.2, instead of working with the group representation ρ, we work with the corresponding algebra homomorphism R : OrGs Ñ M 2 pKq.Write OrρpGqs Ď M 2 pKq for the image of R. Then OrρpGqs is the O-algebra spanned by the image of G in GL 2 pKq Ď M 2 pKq.
Using the same calculation and induction, we deduce the following corollary: Corollary 3.3.Let R : OrGs Ñ M 2 pKq be the algebra homomorphism corresponding to ρ : G Ñ GL 2 pKq.Suppose that there is an integer n ě 1 and O-algebra homomorphisms χ 1 , χ 2 : OrGs Ñ O{π n O such that the characteristic polynomial P ρpgq ptq of ρpgq factors as P ρpgq ptq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q for all g P G. Then P Rpgq ptq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q for all g P OrGs.
We can now prove Theorem 1.2: Proof of Theorem 1.2.If X pρq " X , then ρpGq contains only scalar matrices and the claim is easy.Hence, we may assume that X pρq has boundary points in the sense of Definition 2.19.
The fact that piq implies piiq follows immediately from Lemma 2.3.Indeed, if x, y P X pρq are two points with dpx, yq " n, then there is a lattice Λ x such that Λ x {π n Λ x contains a free, rank one, G-stable O{π n O-submodule.Let χ 1 : G Ñ pO{π n Oq ˆbe the associated representation, and let χ 2 be the quotient of ρ Λx pmod π n q by χ 1 .Then pχ 1 , χ 2 q is a pair of conjugate characters modulo π n .Assume piiq.Let pχ 1 , χ 2 q be a pair of conjugate characters modulo π n .Let g 0 P G be an element for which v π pχ 2 pg 0 q ´χ1 pg 0 qq is minimal.Here, v π is the truncated valuation on O{π n O, taking values 0, 1, . . ., n.
We proceed differently in two cases according to whether X ptρpg 0 quq " X pρq or not.If X ptρpg 0 quq " X pρq, then, by Corollary 2.21 there exist x, y P X ptρpg 0 quq " X pρq with dpx, yq " n, and we are done.
Assume that X ptρpg 0 quq Ľ X pρq.Then there exist two neighbours x, y P X such that x P X pρq and y P X ptρpg 0 quqzX pρq.In particular, there exists an element g 1 P G such that y R X ptρpg 1 quq.By the choice of g 0 , there exists an element a P O such that χ 2 pg 1 q ´χ1 pg 1 q " a pχ 2 pg 0 q ´χ1 pg 0 qq .Then χ 2 pg 1 q ´a ¨χ2 pg 0 q " χ 1 pg 1 q ´a ¨χ1 pg 0 q.Let f " ρpg 1 q ´a ¨ρpg 0 q, which is an element of OrρpGqs, and let δ " χ 1 pg 1 q ´a ¨χ1 pg 0 q.Then, by Lemma 3.1, P f ptq " pt ´δq 2 pmod π n q.
By construction, the point x is a boundary point of X ptf uq.Hence, by Lemma 2.20, there exists a basis pv 1 , v 2 q of Λ x {π n Λ x with respect to which f is represented by the matrix Now, let g P G be any element and let ˜ag b g c g d g ¸P GL 2 pO{π n Oq be the matrix that represents ρpgq with respect to the basis pv 1 , v 2 q.We will show that c g " 0 by computing trpf ¨ρpgqq pmod π n q in two different ways.On the one hand, trpf ¨ρpgqq " tr ˜˜δ 1 0 δ ¸¨˜a g b g c g d g ¸¸" c g `δ ¨trpρpgqq pmod π n q.
Combining the two computations, we see that c g " 0 pmod π n q for all g P G. Thus, the element v 1 generates a free ρpGq-stable O{π n O-submodule of Λ x {π n Λ x .Hence, by Lemma 2.3, there exists a point z P X pρq with dpx, zq " n. □ Remark 3.4.In Theorem 1.2, we can replace the assumption that G is a group with the assumption that G is a set with multiplication, and ρ : G Ñ M 2 pKq is a multiplicative map.Indeed, the notion of conjugate characters and the definition of X pρq apply as they are to the more general case, and the proof of Theorem 1.2 only uses the fact that G is closed under multiplication.
The same arguments as above prove the following slightly more general statement in terms of O-algebras: Theorem 3.5.Let A be an O-algebra and let R : A Ñ M 2 pKq be an O-algebra homomorphism.Assume that A stabilises at least one lattice Λ Ď K 2 .Suppose that there is an integer n ě 1 and O-algebra homomorphisms χ 1 , χ 2 : A Ñ O{π n O such that the characteristic polynomial P Rpgq ptq of Rpgq factors as P Rpgq ptq " pt ´χ1 pgqqpt ´χ2 pgqq pmod π n q for all g P A. Then there exist A-stable lattices

A generalisation of Ribet's Lemma
4.1.Proof of Theorem 1.3.Let s " spχ 1 , χ 2 q.Since ρ is irreducible, by Remark 2.4, the diameter dpρq of X pρq is finite, and by Theorem 1.2 and Lemma 2.3, dpρq " npρq.Let x, y P X pρq be two vertices with dpx, yq " npρq.Then both x and y are leaves of X pρq.Choose lattices Λ x , Λ y representing x, y such that By Lemma 2.3, Λ y {π npρq Λ x is a free, rank 1, G-stable submodule of Λ x {π npρq Λ x , so G acts on it by a character η 1 .Similarly, π npρq Λ x {π npρq Λ y is a G-stable, rank 1, free submodule of Λ y {π npρq Λ y , so G acts on it by a character η 2 .We see that Λ x {π npρq Λ x is an extension of η 2 by η 1 , and Λ y {π npρq Λ y is an extension of η 1 by η 2 .Both extensions are residually non-split because Λ x {πΛ x and Λ y {πΛ y are indecomposable, by Lemma 2.3.
Denote by η 1 and η 2 the reductions of η 1 and η 2 modulo π s .Then Λ x {π s Λ x is a residually non-split extension of η 2 by η 1 and Λ y {π s Λ y is a residually non-split extension of η 1 by η 2 .By Lemma 2.12 we can reorder η 1 , η 2 so that χ i pmod π s q " η i .□ 4.2.Linearly extendable characters and a counterexample.The version of Ribet's Lemma we give in Theorem 1.3 is optimal: if pχ 1 , χ 2 q is a pair of conjugate characters modulo π n , then it is not true in general that there exists a homothety class x P X pρq such that Λ x {π t Λ x is a nonsplit extension of χ 1 by χ 2 when t ą s.When K " Q 2 , one can already find counterexamples arising from elliptic curves over Q.
By Theorem 1.3, there is a lattice for ρ such that Λ{4Λ is a non-split extension of χ cyc by 1 modulo 4. Now, X pρq is exactly the 2-power isogeny graph of E. The lattice Λ should correspond to an isogenous elliptic curve E 1 such that E 1 r2spQq " Z{4Z.One such choice is E 1 : y 2 " x 3 ´x2 `x, with LMFDB label 24.a5.
Since E is not Q-isogenous to any elliptic curve with an 8-torsion point, there is no stable lattice Λ for ρ such that Λ{8Λ contains the trivial representation.Hence, Theorem 1.3 is optimal in this example.See [CLR21, Table 2] for further counterexamples arising from elliptic curves over Q.
More generally, when K is an arbitrary complete valued field, there are two evident obstructions.
First, if such an x P X pρq exists, then χ 1 and χ 2 must factor through the image of ρ.Second, the characters χ 1 , χ 2 must extend O-linearly to OrρpGqs.Indeed, if there is a lattice Λ x such that Λ x {π t Λ x is a non-split extension of χ 1 by χ 2 , then we can choose a basis pv 1 , v 2 q of Λ x such that any g P G is represented by a matrix where r χ 1 pgq, r χ 2 pgq are some lifts of χ 1 pgq, χ 2 pgq to O.Then, with respect to this basis, any element of OrρpGqs is represented by a matrix in Since the maps M 0 pπ t q Ñ O pmod π t q given by `a b cπ t d ˘Þ Ñ a, d pmod π t q are algebra homomorphisms, we see that χ 1 , χ 2 extend O-linearly to OrρpGqs.Definition 4.1.Let χ be a character modulo π n of G.We say that χ is linearly extendable with respect to ρ, or simply linearly extendable, if χ factors through the image of ρ and extends linearly to a map on OrρpGqs.
The character χ is linearly extendable if and only if for each g 1 , . . ., g k P G and a 1 , . . ., a k P O such that Using these obstructions, we prove that Theorem 1.3 is best possible: Proposition 4.2.Fix integers n, s such that n ě 1 and n 2 ď s ď n.Then there exists a group G, a representation ρ : G Ñ GL 2 pOq with npρq " n and a pair of conjugate characters pχ 1 , χ 2 q modulo π n with spχ 1 , χ 2 q " s such that for all t ą spχ 1 , χ 2 q, there does not exist a vertex x P X pρq such that Λ x {π t Λ x is a non-split extension of χ 1 by χ 2 or of χ 2 by χ 1 .and let ρ : G ãÑ GL 2 pOq.By [LM87], any closed subgroup of GL 2 pOq is topologically finitely generated, so G is topologically finitely generated.By the classification of topologically finitely generated abelian profinite groups, the abelianisation G ab of G is isomorphic to Z r p ˆT where T is a finite group.

Proof. Let
We make the assumption that r ě 5, which can always be achieved by choosing K appropriately.For example, for any prime p, we can take K to be any degree 5 extension of Q p .Indeed, G Ą O ˆ, embedded as the diagonal matrices, and O ˆcontains a finite index subgroup isomorphic to O `-Z 5 p as a Z p -module.Let χ 1 , χ 2 be the characters Then pχ 1 , χ 2 q is a pair of conjugate characters of G modulo π n .We have npρq " n.Certainly mpχ 1 , χ 2 q " m, so spχ 1 , χ 2 q " s by definition.The characters χ 1 , χ 2 are evidently linearly extendable.We will construct another pair pη 1 , η 2 q of conjugate characters of G modulo π n such that η 1 pmod π t q is linearly extendable only when t ď spη 1 , η 2 q.Note that, by Remark 2.15, spη 1 , η 2 q " s.
Let g 1 , . . ., g r , . . .g k P G be the pre-images of a minimal set of topological generators for the abelianisation G ab » Z r p ˆT , such that the images of g 1 , . . ., g r generate Z r p .By assumption, r ě 5, and M 2 pKq has dimension 4. Therefore, there exist elements a 1 , . . ., a r P K, not all zero, such that ř r i"1 a i g i " 0 in M 2 pKq, and, up to scaling and reordering, we may assume that a i P O for all i and that a 1 " 1.
Let R n denote the additive group of O{π n O. Let γ P π s R n zπ s`1 R n .Let ε : G Ñ R n be the unique group homomorphism that satisfies εpg 1 q " χ 1 pg 1 q ´1γ, εpg i q " 0 for i " 2, . . ., k.
The image of ε also lies in π s R n , but not in π s`1 R n .
Finally, we show that η 1 pmod π t q is linearly extendable only if t ď s.Recall that ř r i"1 a i g i " 0 where a 1 , . . ., a r P O and a 1 " 1.Using the fact that χ 1 is linearly extendable, we see that, if η 1 pmod π t q is linearly extendable, then 0 " r ÿ i"1 a i η 1 pg i q " r ÿ i"1 a i pη 1 pg i q ´χ1 pg i qq " χ 1 pg 1 qεpg 1 q " γ pmod π t q.
Since γ P π s R n zπ s`1 R n , we see that η 1 pmod π t q is linearly extendable only if t ď s. □

The shape of X pρq
In this section, we give a complete classification of the shape of X pρq in terms of the invariants npρq and mpρq.Conversely, given the shape of X pρq and one additional piece of data, we show how to compute npρq and mpρq.Given reasonable access to the representation ρ (e.g. a finite list of topological generators of ρpGq), it is easy to compute X pρq, but comparatively difficult to compute mpρq and npρq.Throughout this section we assume that the characteristic of the residue field of K is not 2.
After [BC14b, Prop.24], we may assume, without loss of generality, that ρ : G Ñ GL 2 pKq is irreducible.In this case, X pρq is a band whose shape is determined by its radius rpρq and its diameter dpρq [BC14b, Thm.21].
The following theorem, whose proof will occupy the rest of the paper, completes the classification of [BC14b, Thm.45]: Theorem 5.1.Suppose that the residue characteristic of K is not 2. Let ρ : G Ñ GL 2 pKq be an irreducible representation.Then X pρq is a band of diameter dpρq " npρq and radius rpρq " Given mpρq and npρq, Theorem 5.1 gives the exact shape of X pρq.On the other hand, just knowing dpρq and rpρq is not enough to recover npρq and mpρq.By definition, if x, y P X are neighbours, then the generalised ball Bptx, yu, rq of radius r is also a band of diameter 2r `1 and radius r and, in this case, two representations ρ, ρ 1 with npρq " npρ 1 q and mpρq ‰ mpρ 1 q can have X pρq " X pρ 1 q " Bptx, yu, rq.We can tell these cases apart by using thin elements.Definition 5.2.Let g P G.We say that g is a thin element for ρ if X ptρpgquq is an infinite band of radius rpρq.Equivalently, g is a thin element for ρ if there exists a basis of K 2 with respect to which ρpgq is represented by the matrix where α, β P O and v π pβ ´αq " rpρq.
Theorem 5.3.Suppose that the residue characteristic of K is not 2. Let ρ : G Ñ GL 2 pKq be an irreducible representation, and suppose that X pρq is a finite band of diameter dpρq and radius rpρq.Then npρq " dpρq and mpρq " $ ' & ' % dpρq if dpρq " 2rpρq dpρq if dpρq " 2rpρq `1 and if G does not contain a thin element.rpρq otherwise.
Remark 5.4.Suppose that dpρq " 2rpρq `1, i.e. that X pρq is a generalised ball.Then we can determine whether or not G contains a thin element by looking at any list of topological generators of ρpGq, since, in this case, if g, h P G and gh is a thin element, then g or h must be a thin element.Indeed, let pχ 1 , χ 2 q be a pair of conjugate characters modulo π dpρq of G. First, since dpρq ą 2rpρq, by a version of Hensel's lemma (Lemma 5.10), if v π pχ 2 pgq ´χ1 pgqq ď rpρq, for an element g P G, then g is a thin element.Thus, if h and g are not thin elements v π pχ 2 pgq ´χ1 pgqq ą rpρq, and v π pχ 2 phq ´χ1 phqq ą rpρq.
Therefore, if dpρq " 2rpρq `1, then G contains a thin element if and only if one the generators of ρpGq is a thin element.

The invariant kpρq.
Definition 5.5.We denote by kpρq the largest integer k for which the map g Þ Ñ tr ρpgq 2 pmod π k q is multiplicative.
Remark 5.6.Recall that the invariant mpρq is the largest integer m for which there exists a pair of conjugate characters pχ 1 , χ 2 q modulo π npρq such that χ 1 " χ 2 pmod π m q.In particular, mpρq is the largest integer m for which there exists a character χ modulo π m with tr ρpgq " 2χ pmod π m q such that χ lifts to a character modulo π npρq .By contrast, kpρq is the largest integer k for which there exists a character χ modulo π k such that tr ρ " 2χ pmod π k q, but χ need not lift to a character modulo π npρq .
Remark 5.7.There appears to be a gap in the proof of [BC14b,Thm. 45]: in [BC14b, Prop.32], the authors claim that rpρq ď mpρq.However, their proof only shows that rpρq ď kpρq.Indeed, their proof shows that there is a lattice Λ such that the action of G on Λ{π rpρq Λ is by a character, however, it is not clear why this character should extend to a character on Λ{π npρq Λ.
The fact that rpρq ď mpρq is a consequence of the next three lemmas.
Proof.Let x be a point on the nerve of X pρq and let ℓ 1 , ℓ 2 be two leaves of X pρq, both at distance rpρq from x and at distance 2rpρq from one another.Then there exists a basis pv 1 , v 2 q of Λ ℓ 1 such that pv 1 , π rpρq v 2 q is a basis of Λ x and pv 1 , π 2rpρq v 2 q is a basis of Λ ℓ 2 .
For all g P G, let ˜ag b g π 2rpρq c g d g be the matrix representing the action of g with respect to the basis pv 1 , v 2 q.Then the action of g, with respect to the basis pv 1 , π rpρq v 2 q, is ˜ag π rpρq b g π rpρq c g d g

¸.
From the first presentation, we see that the maps g Þ Ñ a g pmod π 2rpρq q and g Þ Ñ d g pmod π 2rpρq q are characters of G.Moreover, since x is in the nerve of X pρq, by definition, Bpx, rpρqq Ď X pρq.Hence, by Lemma 2.7, G acts on Λ x {π rpρq Λ x by a character.Therefore, a g " d g pmod π rpρq q.
Proof.Fix x, y P X pρq with dpx, yq " dpρq.Then there exists a basis pv 1 , v 2 q of Λ x such that pv 1 , π dpρq v 2 q is a basis of Λ y .With respect to this basis, the action of any g P G is represented by a matrix of the form ˜ag b g π dpρq c g d g ¸.Now let z P X pρq be a vertex in the centre of the nerve of X pρq (if the nerve has even cardinality, let z be the central point closest to x).Let ℓ P X pρq be a boundary point with dpℓ, zq " rpρq.If X pρq " BpS, rpρqq and S is a line segment of length a, then dpρq " a `2rpρq and dpx, zq " r `ta{2u.Note that if rpρq " 0, then ℓ " z.
We can choose an element h P G such that both x and ℓ are boundary points of X pthuq.Indeed, let g 1 P G be such that x is a boundary point of X ptρpg 1 quq and let g 2 P G be such that ℓ is a boundary point of X ptρpg 2 quq.Then either ℓ is also a boundary point of X ptρpg 1 quq, x is a boundary point of X ptρpg 2 quq or both x and ℓ are boundary points of X ptρpg 1 g 2 quq.Indeed, if x is not a boundary point of X ptρpg 2 quq, then, by Definition 2.19, all its neighbours are fixed by ρpg 2 q.Since x is a boundary point of X ptρpg 1 quq by definition, there exists a neighbour x 1 of x such that ρpg 1 qΛ x 1 ‰ Λ x 1 , where Λ x 1 is any representative of x 1 .But then ρpg 1 g 2 qΛ x 1 " ρpg 1 qΛ x 1 ‰ Λ x 1 , so x is not a boundary point of X ptρpg 1 g 2 quq.The argument for ℓ is similar.
The vectors pv 1 , π dpx,zq v 2 q form a basis of Λ z , with respect to which, the action of h is represented by the matrix ˜ah π dpx,zq b h π dpρq´dpx,zq c h d h ¸.
Note that, by the definition of z, rpρq ă dpρq´1 2 ď dpx, zq and that rpρq ă dpρq ´dpx, zq.Therefore, modulo π rpρq`1 , h is represented by the matrix ˜ah 0 0 d h ¸pmod π rpρq`1 q.Now, X pρq contains the ball of radius rpρq around z, so v π pd h ´ah q ě rpρq.On the other hand, by the way we chose h, Bpz, rpρq `1q Ď X pthuq, so, by Lemma 2.7, h does not act as a scalar on Λ z {π rpρq`1 Λ z .Hence, v π pd h ´ah q " rpρq.By Lemma 5.10 we see that h is a thin element for ρ. □ Proof of Theorems 5.1 and 5.3.By Theorem 1.2, we always have dpρq " npρq.Note that Theorem 5.3 implies Theorem 5.1, so we only need to prove the former theorem.
Now suppose that G does not contain a thin element.Suppose for contradiction that mpρq ‰ npρq.Then, by Lemma 5.8, 2mpρq ă npρq.Thus, there exists an element g P G such that the characteristic polynomial of ρpgq factors as P ρpgq ptq " pt ´αqpt ´βq pmod π npρq q, with α, β P O{π npρq O satisfying v π pβ ´αq " mpρq ă npρq{2.Thus, by Lemma 5.10, g is a thin element in G, a contradiction.□

0Figure 1 . 0 Figure 2 .Figure 3 .
Figure 1.A ball of radius 3 in the Bruhat-Tits tree for Q 3 local field, ρ is continuous and χ : G Ñ pO{π n Oq îs a character that factors through the image of ρ, then χ is automatically continuous.Indeed, any finite index subgroup of the closed subgroup ρpGq Ď GL 2 pOq is open in ρpGq [LM87, Thm.A].