On the reductions of certain two-dimensional crystalline representations, III
aa r X i v : . [ m a t h . N T ] F e b ON THE REDUCTIONS OF CERTAINTWO-DIMENSIONAL CRYSTALLINEREPRESENTATIONS, III
BODAN ARSOVSKI
Abstract.
A conjecture of Breuil, Buzzard, and Emerton says that the slopesof certain reducible p -adic Galois representations must be integers. In [Ars] weshowed this conjecture for representations that lie over certain “non-subtle”components of weight space. This article is a continuation of [Ars] in whichwe also show the conjecture for the “subtle” components for slopes less than p − . Introduction and results
Background.
Let p be an odd prime number and k > a be an element of Z p such that v p ( a ) >
0. Let us denote ν = ⌊ v p ( a ) ⌋ + 1 ∈ Z > .With this data one can associate a certain two-dimensional crystalline p -adic rep-resentation V k,a with Hodge–Tate weights (0 , k − V k,a as the semi-simplificationof the reduction modulo the maximal ideal m of Z p of a Galois stable Z p -latticein V k,a (with the resulting representation being independent of the choice of lat-tice). The question of computing V k,a has been studied extensively, and we referto the introduction of [Ars] for a brief exposition of it. Partial results have beenobtained by Fontaine, Edixhoven, Breuil, Berger, Li, Zhu, Buzzard, Gee, Bhat-tacharya, Ganguli, Ghate, et al (see [Ber10], [Bre03a], [Bre03b], [Edi92], [BLZ04],[BG15], [BG09], [BG13], [GG15]). A conjecture of Breuil, Buzzard, and Emertonsays the following. Conjecture A. If k is even and v p ( a ) Z then V k,a is irreducible. The main result of [Ars] is that this conjecture is true over certain “non-subtle”components of weight space. We say that a weight k belongs to a “non-subtle”component of weight space if and only if k , , . . . , ν, ν + 1 mod p − . Thus there are max { p − − ν + 1 , } many “non-subtle” components of weight space.This article is a continuation of [Ars] in which we also show the conjecture for the“subtle” components for slopes less than p − . The main result we show is thefollowing theorem. Theorem 1.
Conjecture A is true when the slope is less than p − . Date : February 2021. Computing V k,a by computing Θ k,a From now on we assume the notation from sections 2, 3, 4, and 6 of [Ars]. Moreover,we assume that k > p as in section 5 of [Ars]. Theorem 2 in [Ars] implies thatour main theorems can be rewritten in the following equivalent forms. Recall thatwe assume p >
Theorem 2. If k ∈ Z and v p ( a ) ∈ (0 , p − ) \ Z then Θ k,a is irreducible. Thus our task is to prove theore 2.3.
Combinatorics
Throughout the proof we will refer to the combinatorial results in section 8 of [Ars].For convenience, we reproduce the statements here in the form we will use.
Lemma 3.
Suppose throughout this lemma that n, t, y ∈ Z , b, d, k, l, w ∈ Z > , m, u, v ∈ Z > . (1) If u ≡ v mod ( p − p m − then M u,n ≡ M v,n mod p m . ( c-a ) (2) Suppose that u = t u ( p −
1) + s u with s u = u , so that s u ∈ { , . . . , p − } and t u ∈ Z > . Then M u = 1 + [ u ≡ p −
0] + t u s u p + O (cid:0) t u p (cid:1) . ( c-b ) (3) If n then M u,n = P − ni =0 ( − i (cid:0) − ni (cid:1) M u − n − i, . ( c-c ) (4) If n > then M u,n ≡ (1 + [ u ≡ p − n ≡ p − (cid:0) un (cid:1) mod p. ( c-d ) (5) If u > ( b + l ) d and l > w then P j ( − j − b (cid:0) lj − b (cid:1)(cid:0) u − djw (cid:1) = [ w = l ] d l . ( c-e ) (6) If X is a formal variable then (cid:0) Xt + l (cid:1)(cid:0) tw (cid:1) = P v ( − w − v (cid:0) l + w − v − w − v (cid:1)(cid:0) Xv (cid:1)(cid:0) X − vt + l − v (cid:1) . ( c-f ) Consequently, if b + l > d + w then P i (cid:0) b − d + li ( p − l (cid:1)(cid:0) i ( p − w (cid:1) = P v ( − w − v (cid:0) l + w − v − w − v (cid:1)(cid:0) b − d + lv (cid:1) M b − d + l − v,l − v . ( c-g ) (7) We have P j ( − j (cid:0) yj (cid:1)(cid:0) y + l − jw − j (cid:1) = ( − w (cid:0) w − l − w (cid:1) . ( c-i ) (8) We have P j (cid:0) u − j − (cid:1)(cid:0) − lj − w (cid:1) = ( − u − w (cid:0) l − wu − w (cid:1) . ( c-j ) (9) We have P j ( − j (cid:0) jb (cid:1)(cid:0) lj − w (cid:1) = ( − l + w (cid:0) wl + w − b (cid:1) . ( c-k ) EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 3
Lemma 4.
Let α ∈ Z ∩ [0 , . . . , rp +1 ] and let { D i } i ∈ Z be a family of elements of Z p such that D i = 0 for i [0 , r − αp − ] and ϑ w ( D • ) = 0 for all w < α . Then P i D i x i ( p − α y r − i ( p − − α = θ α h for some polynomial h with integer coefficients. Lemma 5.
For α, λ, µ ∈ Z > let L α ( λ, µ ) be the ( α + 1) × ( α + 1) matrix with entries L l,j = P αk =0 j ! l ! (cid:0) µλ (cid:1) k s ( l, k ) s ( k, j ) , where s ( l, k ) are the Stirling numbers of the first kind and s ( k, j ) are the Stirlingnumbers of the second kind. Then L α ( λ, µ ) (cid:16)(cid:0) λX (cid:1) , . . . , (cid:0) λXα (cid:1)(cid:17) T = (cid:16)(cid:0) µX (cid:1) , . . . , (cid:0) µXα (cid:1)(cid:17) T . Lemma 6.
For α ∈ Z > let B α be the ( α + 1) × ( α + 1) matrix with entries B i,j = j ! P αk,l =0 ( − i + l + k l ! (cid:0) li (cid:1) (1 − p ) − k s ( l, k ) s ( k, j ) , where s ( i, j ) and s ( k, j ) are the Stirling numbers of the first and second kind,respectively. Let { X i,j } i,j > be formal variables. For β ∈ Z > such that α > β let S ( α, β ) = ( S ( α, β ) w,j ) w,j α be the ( α + 1) × ( α + 1) matrix with entries S ( α, β ) w,j = P βi =1 X i,j (cid:0) i ( p − w (cid:1) . Then B α S ( α, β ) is zero outside the rows indexed , . . . , β and ( B α S ( α, β )) i,j = X i,j for i ∈ { , . . . , β } . Lemma 7.
For u, v, c ∈ Z let us define F u,v,c ( X ) = P w ( − w − c (cid:0) wc (cid:1)(cid:0) Xw (cid:1) ∂ (cid:0) X + u − wv − w (cid:1) ∈ Q p [ X ] . Then F u,v,c ( X ) = (cid:0) uv − c (cid:1)(cid:0) Xc (cid:1) ∂ − (cid:0) uv − c (cid:1) ∂ (cid:0) Xc (cid:1) . Lemma 8.
Let X and Y denote formal variables, and let c j = ( − j α ! (cid:16) X + j +1 j +1 (cid:0) Yα − j − (cid:1) + (cid:0) Yα − j (cid:1)(cid:17) ∈ Q [ X, Y ] ⊂ Q ( X, Y ) be polynomials over Q of degrees α − j , for j α . Let M = ( M w,j ) w,j α be the ( α + 1) × ( α + 1) matrix over Q ( X, Y ) with entries M w, = ( − w ( Y − X ) X w Y w +1 ,M w,j = P v ( − w − v (cid:0) j + w − v − w − v (cid:1)(cid:0) X + jv (cid:1) (cid:16)(cid:0) Y + j − vj − v (cid:1) − (cid:0) X + j − vj − v (cid:1)(cid:17) , for w α and < j α . Then the first α − entries of M c = M ( Y α , c , . . . , c α ) T = ( d , . . . , d α ) T are zero, and d α = ( Y − X ) α +1 Y − α . BODAN ARSOVSKI
Lemma 9.
Suppose that s, α, β ∈ Z are such that β α s − p − . Let B = B α denote the matrix defined in lemma 6. Let M denote the ( α + 1) × ( α + 1) matrix with entries in F p such that if i ∈ { , . . . , β } and j ∈ { , . . . , α } then M i,j = (cid:0) βi (cid:1) · ( (cid:0) s − α − β + ii (cid:1) − ( − i +1 if j = 0 , (cid:0) s − α − β + jj − i (cid:1) if j > , and if i ∈ { , . . . , α }\{ , . . . , β } and j ∈ { , . . . , α } then M i,j is the reduction mod-ulo p of p − [ j =0] P αw =0 B i,w P v ( − w − v (cid:0) j + w − v − w − v (cid:1)(cid:0) s + β ( p − − α + jv (cid:1) ∂ · P βu =0 (cid:0) s + β ( p − − α + j − vu ( p − j − v (cid:1) − [ i = 0] p − [ j =0] (cid:0) s + β ( p − − α + jj (cid:1) ∂ − [ j = 0] P αw =0 B i,w ( − w (cid:0) s + β ( p − − αw (cid:1) w !( s − α ) w +1 . Then there is a solution of M ( z , . . . , z α ) T = (1 , , . . . , T such that z = 0 . Now let us prove some additional combinatorial results.
Lemma 10.
Suppose that s, α ∈ Z > are such that s ∈ { , , . . . , p − } and s α < s and α p − . For w, j ∈ Z > let F w,j ( z ) ∈ F p [ z ] denote the polynomial P v ( − w − v (cid:0) j + w − v − w − v (cid:1)(cid:0) z − α + jv (cid:1)(cid:0) s − α + j − vj − v (cid:1) − (cid:0) z − α + jj (cid:1)(cid:0) w (cid:1) − (cid:0) z − α + js − α (cid:1)(cid:0) z − sw (cid:1) . Let C ( z ) , . . . , C α ( z ) ∈ F p [ z ] denote the polynomials C j ( z ) = (cid:0) αs − α − (cid:1) − s − zα +1 if j = 0 , ( − j +1 j +1 (cid:0) s − α − α − j (cid:1) ( z − α ) if j ∈ { , . . . , α } . Let F ( z ) , F ( z ) ∈ F p [ z ] denote the polynomials F ( z ) = P αj =0 C j ( z ) F w,j ( z ) ,F ( z ) = − [ w = 0] (cid:0) s − z − α +1 (cid:1) . Note that all of these polynomials depend on s and α . Then F ( z ) = F ( z ) .Proof. Let us first show that C ( z ) (cid:0) z − αs − α (cid:1) + P αj =1 C j ( z ) (cid:0) z − α + js − α (cid:1) = ( − α +1 ( z − α ) s − α . (1)Since C ( z ) (cid:0) z − αs − α (cid:1) = (cid:0) αs − α − (cid:1) − s − zα +1 z − αs − α (cid:0) z − α − s − α − (cid:1) , this is equivalent to (cid:0) αs − α − (cid:1) − s − zα +1 (cid:0) z − α − s − α − (cid:1) + P αj =1 ( − j +1 ( α − j +1) j +1 (cid:0) s − αα − j +1 (cid:1)(cid:0) z − α + js − α (cid:1) = ( − α +1 . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 5
The polynomial on the left side has degree at most s − α . The coefficient of z s − α in it is − (2 α − s +1)!( α +1)! plus s − α − P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1) = s − α − P j ( − j +1 j +1 [ X s − α + j − ](1 + X ) s − α − = P j ( − j +1 ( j +1) [ X j ] X α − s − (1 + X ) s − α − = s − α − R − Y α − s +1 (1 + Y ) s − α − d Y = ( − s (2 α − s +1)!( α +1)! . Since s is even, that coefficient is zero. Therefore it is enough to show that the twopolynomials are equal when evaluated at z ∈ { α + 1 , . . . , s } . At these points thepolynomial on the left side is equal to( s − α ) P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) s − α − γ + js − α (cid:1) for γ ∈ { , . . . , s − α − } . We have P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) s − α − γ + js − α (cid:1) = P j ( − s − α + j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) γ − j − s − α (cid:1) = P u (cid:0) γu (cid:1) P j ( − s − α + j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) − j − s − α − u (cid:1) = P u (cid:0) γu (cid:1) P j ( − s − α + j s − α − u (cid:0) s − α − α − j (cid:1)(cid:0) − j − s − α − u − (cid:1) = P u ( − u +1 s − α − u (cid:0) γu (cid:1) P j (cid:0) s − α − α − j (cid:1)(cid:0) − s + α + uj +2 (cid:1) = P u ( − u +1 s − α − u (cid:0) γu (cid:1)(cid:0) u − α +2 (cid:1) = ( − α +1 s − α . The third equality follows from (cid:0) γu (cid:1) = 0 for u > s − α −
1, and the last equalityfollows from (cid:0) u − α +2 (cid:1) = 0 for u ∈ { , . . . , s − α − } . In particular, (1) is indeed true.So both F ( z ) and F ( z ) have degree at most α + 1, and therefore they are equalif they are equal when evaluated at z ∈ { s + γ ( p − | γ ∈ { , . . . , α + 1 }} . It is easy to verify that F ( s ) = F ( s ), and when z ∈ { s + γ ( p − | γ ∈ { , . . . , α + 1 }} the fact that P γ − i =1 (cid:0) s + γ ( p − − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = F w,j ( s + γ ( p − c-g )) implies that the equation F ( s + γ ( p − F ( s + γ ( p − P αj =0 C j ( s + γ ( p − P γ − i =1 (cid:0) s + γ ( p − − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = − [ w = 0] (cid:0) − γ ( p − − α +1 (cid:1) . Note that (cid:0) − γ ( p − − α +1 (cid:1) = (cid:0) γ − α +1 (cid:1) = 0 and therefore the right side vanishes. Letus reiterate that all computations done in this proof are over F p . Let us write C γj = C j ( s + γ ( p − P αj =0 C γj P γ − i =1 (cid:0) s + γ ( p − − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = 0follows if P αj =0 C γj (cid:0) s + γ ( p − − α + ji ( p − j (cid:1) = 0for all i ∈ { , . . . , γ − } . If j > C γj = 0 then j > α − s + 1 > α + γ − s BODAN ARSOVSKI and consequently (cid:0) s + γ ( p − − α + ji ( p − j (cid:1) = ( (cid:0) γi (cid:1)(cid:0) s − α − γ + js − α − γ + i (cid:1) if s − α − γ + i > (cid:0) γi − (cid:1)(cid:0) s − α − γ + jp + s − α − γ + i (cid:1) if s − α − γ + i < (cid:0) γi (cid:1)(cid:0) s − α − γ + js − α − γ + i (cid:1) . On the other hand, (cid:0) s + γ ( p − − αi ( p − (cid:1) = (cid:0) γ − i − (cid:1)(cid:0) s − α − γs − α − γ + i (cid:1) . Since (cid:0) γ − i − (cid:1) = iγ (cid:0) γi (cid:1) ∈ F × p (as that 0 < i < γ α + 1), what we want to show is that C γ iγ (cid:0) s − α − γs − α − γ + i (cid:1) + P αj =1 C γj (cid:0) s − α − γ + js − α − γ + i (cid:1) = 0for all i ∈ { , . . . , γ − } . That is equivalent to F ( s + γ ( p − , where F ( z ) ∈ F p [ z ] is defined as F ( z ) = (cid:0) αs − α − (cid:1) − s − z − wα +1 (cid:0) z − α − s − α − w − (cid:1) + P αj =1 ( − j +1 ( s − α − w ) j +1 (cid:0) s − α − α − j (cid:1)(cid:0) z − α + js − α − w (cid:1) with w = γ − i >
0. The degree of F ( z ) is at most s − α − w , and in fact thecoefficient of z s − α − w in it is − ( s − α − w (2 α − s +1)!( α +1)! plus s − α − w − P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1) = ( s − α − w (2 α − s +1)!( α +1)! , i.e. the coefficient of z s − α − w in it is zero. Therefore the degree of F ( z ) is less than s − α − w , so it is enough to show that F ( z ) is equal to zero when evaluated at z ∈ { α + 1 , . . . , s − w } . At these points F ( z ) is equal to( s − α − w ) P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) s − α − γ + js − α − w (cid:1) for γ ∈ { w, . . . , s − α − } . We have P j ( − j +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) s − α − γ + js − α − w (cid:1) = P j ( − s − α + j − w +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) γ − j − w − s − α − w (cid:1) = P u (cid:0) γ − wu (cid:1) P j ( − s − α + j − w +1 j +1 (cid:0) s − α − α − j (cid:1)(cid:0) − j − s − α − u − w (cid:1) = P u (cid:0) γ − wu (cid:1) P j ( − s − α + j − w s − α − u − w (cid:0) s − α − α − j (cid:1)(cid:0) − j − s − α − u − w − (cid:1) = P u ( − u +1 s − α − u − w (cid:0) γ − wu (cid:1) P j (cid:0) s − α − α − j (cid:1)(cid:0) − s + α + u + wj +2 (cid:1) = P u ( − u +1 s − α − u − w (cid:0) γ − wu (cid:1) P j (cid:0) u + w − α +2 (cid:1) = 0 . The last equality follows from (cid:0) γ − wu (cid:1) = 0 for u
6∈ { , . . . , s − α − w − } . This proves that indeed F ( z ) = 0 and therefore that F ( z ) = F ( z ). Lemma 11.
Suppose that α ∈ Z > . For w, j ∈ { , . . . , α } let F w,j ( z, ψ ) ∈ F p [ z, ψ ] denote the polynomial P v ( − w − v (cid:0) j + w − v − w − v (cid:1)(cid:0) z − α + jv (cid:1) (cid:16)(cid:0) ψ − α + j − vj − v (cid:1) − (cid:0) z − α + j − vj − v (cid:1)(cid:17) . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 7
Note that this depends on α . Then P αj =1 ( − α − j (cid:0) ψ − α +1 α − j (cid:1) F w,j ( z, ψ ) = ( − α ([ w = α ] − [ w = 0]) (cid:0) ψ − zα (cid:1) . Proof.
Both sides of the equation we want to prove have degree α and thecoefficient of z α on each side is α ! ([ w = α ] − [ w = 0]). So the two sides are equalif they are equal when evaluated at the points ( z, ψ ) such that( z, ψ ) ∈ { ( u + γ ( p −
1) + α, u + α ) | u ∈ { , . . . , α } , γ ∈ { , . . . , α − }} . The right side is zero when evaluated at these points, and F w,j ( u + γ ( p −
1) + α, u + α ) = P γi =1 (cid:0) u + γ ( p − ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) by ( c-g ). Thus we want to show that P αj =1 ( − α − j (cid:0) u +1 α − j (cid:1) P γi =1 (cid:0) u + γ ( p − ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = O ( p )for 0 u, w α and 0 γ < α . Since (cid:0) u + γ ( p − ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = (cid:0) γi (cid:1)(cid:0) u + j − γj − i (cid:1)(cid:0) − iw (cid:1) + O ( p ) , that is equivalent to P i,j> ( − α + w − i (cid:0) u +1 α − j (cid:1)(cid:0) γi (cid:1)(cid:0) γ − u − i − j − i (cid:1)(cid:0) i + w − w (cid:1) = O ( p ) . This follows from the facts that P j> (cid:0) u +1 α − j (cid:1)(cid:0) γ − u − i − j − i (cid:1) = (cid:0) γ − iα − i (cid:1) for i > (cid:0) γi (cid:1)(cid:0) γ − iα − i (cid:1) = (cid:0) αi (cid:1)(cid:0) γα (cid:1) = 0since γ ∈ { , . . . , α − } . Lemma 12.
Suppose that s, α, β ∈ Z are such that s ∈ { , , . . . , p − } and α = β = s + 1 . Let M denote the ( α + 1) × ( α + 1) matrix with entries in F p defined in lemma 14.Suppose that C , . . . , C α ∈ F p are defined as C j = ( − α + j +1 α (cid:0) α − j − (cid:1) . Then M ( C , . . . , C α ) T = (0 , . . . , , T . Proof.
The equation associated with the i th row of M is straightforward if i
6∈ { , α } . Since M ,j is equal to P αl,v =0 ( − l + v (cid:0) j − vl − v (cid:1)(cid:0) j − v (cid:1) ∂ (cid:0) α + j − v − j − v (cid:1) (1 + [ α = 2 & j = v ]) − (cid:0) j − j (cid:1) ∂ − (cid:0) j − α − (cid:1)(cid:0) α (cid:1) ∂ BODAN ARSOVSKI and since P j ( − j (cid:0) α − j − (cid:1) j ( j − = α , P j ( − j (cid:0) α − j − (cid:1)(cid:0) j − s − α (cid:1)(cid:0) α (cid:1) ∂ = ( − α (cid:0) α (cid:1) ∂ = − α , P αl ( − l + α (cid:0) − αl − α (cid:1)(cid:0) α (cid:1) ∂ (cid:0) p − α − (cid:1) = − α , P αl ( − l + α (cid:0) − αl − α (cid:1)(cid:0) α (cid:1) ∂ (cid:0) α − (cid:1) = − [ α =2] α , the equation associated with the zeroth row is P j ( − j (cid:0) α − j − (cid:1) P αl,v =0 ( − l + v (cid:0) j − vl − v (cid:1)(cid:0) j − v (cid:1) ∂ (cid:0) α + j − v − j − v (cid:1) = − α , and it follows from the fact that P αl =0 ( − l (cid:0) li (cid:1)(cid:0) j − vl − v (cid:1) = ( − j (cid:0) vj − i (cid:1) for 0 v j α . This shows the equation associated with the zeroth row. Since M α,j is equal to[ j = 2] (cid:0) α (cid:1) ∂ + [ j = α ] F α,α, ( α − − (cid:0) j − α − (cid:1) ∂ − ( − α (cid:0) j − α − (cid:1)(cid:0) − α (cid:1) ∂ , the equation associated with the α th row is (cid:0) α (cid:1) ∂ − P j ( − j (cid:0) α − j − (cid:1)(cid:0) j − α − (cid:1) ∂ = (cid:0) − α (cid:1) ∂ − ( − α F α,α, ( α −
2) + ( − α +1 α , and it follows from the facts that F α,α, ( α −
2) = F α,α, ( −
1) = ( − α (cid:0) − α (cid:1) ∂ and that the polynomial (cid:0) Xα − (cid:1) ∂ ∈ F p [ X ] has degree less than α − α = 2) and therefore P j ( − j (cid:0) α − j − (cid:1)(cid:0) j − α − (cid:1) ∂ = 0 . This shows the equation associated with the α th row and concludes the proof. Lemma 13.
Suppose that s, α, β ∈ Z are such that s ∈ { , , . . . , p − } and α = s + 1 and β ∈ { s , s + 1 } . Let A denote the β × β matrix with entries in Q p defined as A = (cid:16) p [ j =1] − [ i β − α +1] (cid:0) s + β ( p − − α + ji ( p − j (cid:1)(cid:17) i<β, α − β
3) submatrix of A is upper triangularwith units on the diagonal. Moreover, since P j ( − j (cid:0) s − β − j − α − i − j − (cid:1)(cid:0) α − j (cid:1) = P j ( − j (cid:0) β − j − i − (cid:1)(cid:0) β − j (cid:1) = 0 , P j ( − j − ( j − (cid:0) s − β − jα − i − j (cid:1)(cid:0) α − j (cid:1) = P j ( − j − ( j − (cid:0) β − ji − (cid:1)(cid:0) βj (cid:1) = 0 , EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 9 all but the top two entries of each of the vectors c α − − (cid:0) α − (cid:1) c α − + · · · + ( − α − (cid:0) α − α − (cid:1) c , c α − (cid:0) α − (cid:1) c α − + · · · + ( − α − ( α − (cid:0) α − α − (cid:1) c are zero. Thus it is enough to show that the 2 × F p ). This 2 × (cid:16) e , e , ( − β β ( − β β ( β − (cid:17) with e , = β P β − j =0 ( − j (cid:0) β − j (cid:1)(cid:0) β − j − β − j (cid:1) ∂ = P β − j =0 ( − j ββ − j (cid:0) β − j (cid:1) = P β − j =0 ( − j (cid:0) βj (cid:1) = ( − β +1 ,e , = β P βj =0 ( − j − ( j − (cid:0) βj (cid:1)(cid:0) β − jβ − j +1 (cid:1) ∂ = P βj =0 ( − j − β ( j − β − j +1 (cid:0) βj (cid:1) = P βj =0 ( − j − β ( j − β +1 (cid:0) β +1 j (cid:1) = ( − β − β β +1 , so it has determinant ββ +1 ∈ F × p . Now suppose that β = α and denote the columnsof A by c , . . . , c α . The bottom left ( α − × ( α −
1) submatrix of A is uppertriangular with units on the diagonal, all but the top entry of the vector c α − (cid:0) α − (cid:1) c α − + · · · + ( − α − (cid:0) α − α − (cid:1) c are zero, and that top entry is β P β − j =0 ( − j (cid:0) β − j (cid:1)(cid:0) β − j − β − j (cid:1) ∂ = P β − j =0 ( − j β − (cid:0) βj (cid:1) = ( − β ∈ F × p . Therefore A is invertible. Lemma 14.
Suppose that s, α, β ∈ Z are such that s ∈ { , , . . . , p − } and s α s and β α. Let M denote the ( α + 1) × ( α + 1) matrix with entries in F p such that if i ∈ { , . . . , β − } and j ∈ { , . . . , α } then M i,j = (cid:0) s + β ( p − − α + ji ( p − j (cid:1) , and if i ∈ { , . . . , α }\{ , . . . , β − } and j ∈ { , . . . , α } then M i,j = P αl,v =0 ( − i + l + v (cid:0) li (cid:1)(cid:0) j − vl − v (cid:1)(cid:0) s − α − β + jv (cid:1) ∂ (cid:0) s − α + j − vj − v (cid:1) − [ i = 0] (cid:0) s − α − β + jj (cid:1) ∂ − [ i = β ] (cid:0) s − α − β + js − α (cid:1) ∂ − ( − i (cid:0) s − α − β + js − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) l − β − l (cid:1) ∂ . Suppose that C , . . . , C α ∈ F p are defined as C j = ( if j = 0 , ( − j +1 ( s − α − β ) β (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1) if j ∈ { , . . . , α } . Then M ( C , C , . . . , C α ) T = (cid:18) ( − α + β +1 ( s − α )( α − β +1) β (2 α − s +1) ( αβ ) (cid:0) αs − α (cid:1) , , . . . , (cid:19) T . Proof.
Let us denote the rows of M by r , . . . , r α . Note that if j > C j = 0 then j > α − s , so s − α + j > α and in particular (cid:0) s − α + j − vj − v (cid:1) = (cid:0) s − α + j − vj − v (cid:1) . We have the following string of equations: P j > ( − j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α − β + js − α (cid:1) = P u (cid:0) βu (cid:1) P j > ( − s − α + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) − j − s − α − u (cid:1) = P u (cid:0) βu (cid:1)(cid:0) α − u +1 s − α − u (cid:1) P j > ( − j + u +1 (cid:0) α +1 j +1 (cid:1)(cid:0) s − α + j − uα − u +1 (cid:1) = P u (cid:0) βu (cid:1)(cid:0) α − u +1 s − α − u (cid:1) (cid:16) ( − u +1 (cid:0) s − α − u − α − u +1 (cid:1) + P j ( − j + u +1 (cid:0) α +1 j +1 (cid:1)(cid:0) s − α + j − uα − u +1 (cid:1)(cid:17) = P u (cid:0) βu (cid:1)(cid:0) α − u +1 s − α − u (cid:1) (cid:16) ( − u +1 (cid:0) s − α − u − α − u +1 (cid:1) + [ u = 0]( − α +1 (cid:17) = ( − α (cid:16)(cid:0) βs − α (cid:1) − (cid:0) α +1 s − α (cid:1)(cid:17) . The first two equalities amount to rewriting the binomial coefficients. The thirdequality amounts to computing the inner sum. The fourth equality follows from ( c-e ).The fifth equality amounts to computing the outer sum. This string of equationsimplies that P αj =0 C j (cid:0) s − α − β + js − α (cid:1) = ( − α +1 s − α − ββ (cid:0) α +1 s − α (cid:1) . Our task is to compute r i ( C , C , . . . , C α ) T for i ∈ { , . . . , α } . • Computing r ( C , C , . . . , C α ) T . If j > α − s then P αl =0 ( − l (cid:0) li (cid:1)(cid:0) j − vl − v (cid:1) = ( − j (cid:0) vj − i (cid:1) for 0 v j α and therefore M ,j = P αv =0 ( − j + v (cid:0) vj (cid:1)(cid:0) s − α − β + jv (cid:1) ∂ (cid:0) s − α + j − vj − v (cid:1) − (cid:0) s − α − β + jj (cid:1) ∂ − (cid:0) s − α − β + js − α (cid:1) P αl =0 (cid:0) l − β − l (cid:1) ∂ = − (cid:0) s − α − β + js − α (cid:1) P αl =0 (cid:0) l − β − l (cid:1) ∂ . The second equality follows from the fact that (cid:0) vj (cid:1) = 0 if v < j . We also have M , = P αl,v =0 ( − l + v (cid:0) − vl − v (cid:1)(cid:0) s − α − βv (cid:1) ∂ (cid:0) s − α − v − v (cid:1) − (cid:0) s − α − βs − α (cid:1) P αl =0 (cid:0) l − β − l (cid:1) ∂ = P αl,v =0 ( − α + l + v (cid:0) − vl − v (cid:1)(cid:0) s − α − βv (cid:1) ∂ (cid:0) vs − α (cid:1) − (cid:0) s − α − βs − α (cid:1) P αl =0 (cid:0) l − β − l (cid:1) ∂ = P αl,v =0 ( − α + l + v (cid:0) − vl − v (cid:1)(cid:0) s − α − βv (cid:1) ∂ (cid:0) vs − α (cid:1) + (cid:0) s − α − βs − α (cid:1) P αl =0 ( − l P v ( − v +1 (cid:0) s − α − βv (cid:1) ∂ (cid:0) s − α − vl − v (cid:1) . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 11
The third equality follows from lemma 7. Thus r ( C , . . . , C α ) T is equal to P αl,v =0 ( − α + l + v (cid:0) s − α − βv (cid:1) ∂ (cid:16)(cid:0) − vl − v (cid:1)(cid:0) vs − α (cid:1) + s − α − ββ (cid:0) α +1 s − α (cid:1)(cid:0) s − α − vl − v (cid:1)(cid:17) = ( − α P αv =0 (cid:0) s − α − βv (cid:1) ∂ (cid:16)(cid:0) αv (cid:1)(cid:0) vs − α (cid:1) + s − α − ββ (cid:0) α +1 s − α (cid:1)(cid:0) α − sα − v (cid:1)(cid:17) = (cid:16)(cid:0) αs − α (cid:1) + s − α − ββ (cid:0) α +1 s − α (cid:1)(cid:17) P αv =0 ( − v (cid:0) s − α − βv (cid:1) ∂ (cid:0) s − α − v − α − v (cid:1) = ( − α + β +1 β ( αβ ) (cid:16)(cid:0) αs − α (cid:1) + s − α − ββ (cid:0) α +1 s − α (cid:1)(cid:17) = ( − α + β +1 ( s − α )( α − β +1) β (2 α − s +1) ( αβ ) (cid:0) αs − α (cid:1) . The third equality follows from lemma 7. Thus we have computed r ( C , . . . , C α ) T = ( − α + β +1 ( s − α )( α − β +1) β (2 α − s +1) ( αβ ) (cid:0) αs − α (cid:1) . • Computing r i ( C , C , . . . , C α ) T for i ∈ { , . . . , β − } . Let w ∈ Z be such that i = β − w ∈ { , . . . , β − } . Then P j > − j +1 ( s − α − β ) β (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α − β + js − α − w (cid:1) = P u (cid:0) α − β +1 u (cid:1) P j > − j +1 ( s − α − β ) β (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α + j − s − α − w − u (cid:1) = P u (cid:0) α − β +1 u (cid:1)(cid:0) α − w − u +1 s − α − w − u (cid:1) P j > − j +1 ( s − α − β ) β (cid:0) jα − w − u +1 (cid:1)(cid:0) α +1 j +1 (cid:1) = ( − α − w − u s − α − ββ P u (cid:0) α − β +1 u (cid:1)(cid:0) α − w − u +1 s − α − w − u (cid:1) = s − α − ββ P u (cid:0) α − β +1 u (cid:1)(cid:0) s − α − s − α − w − u (cid:1) = s − α − ββ (cid:0) s − α − β − s − α − w (cid:1) = − iβ (cid:0) s − α − βs − α − w (cid:1) . The third equality follows from ( c-e ). Consequently, if i ∈ { , . . . , β − } then r i ( C , . . . , C α ) T = 0 . • Computing r i ( C , C , . . . , C α ) T for i ∈ { β, . . . , α } . For these i we have M i, = P αl,v =0 ( − α + i + l + v (cid:0) li (cid:1)(cid:0) vs − α (cid:1)(cid:0) s − α − βv (cid:1) ∂ (cid:0) − vl − v (cid:1) − [ i = β ] (cid:0) s − α − βs − α (cid:1) ∂ − ( − i (cid:0) s − α − βs − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) l − β − l (cid:1) ∂ , and for j > α − s we also have M i,j = P αl,v =0 ( − i + l + v (cid:0) li (cid:1)(cid:0) j − vl − v (cid:1)(cid:0) s − α − β + jv (cid:1) ∂ (cid:0) s − α + j − vj − v (cid:1) − [ i = β ] (cid:0) s − α − β + js − α (cid:1) ∂ − ( − i (cid:0) s − α − β + js − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) l − β − l (cid:1) ∂ . The identity P αj =0 ( − j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) z + s − α + js − α (cid:1) ∂ = ∂∂z (cid:16)P αj =0 ( − j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) z + s − α + js − α (cid:1)(cid:17) = ∂∂z (cid:16)(cid:0) z + s − α − s − α (cid:1) − (cid:0) s − α − s − α (cid:1)(cid:17) = (cid:0) z + s − α − s − α (cid:1) ∂ is true over Q p [ z ]. By evaluating at z = − β we get P αj =1 C j (cid:0) s − α − β + js − α (cid:1) ∂ = s − α − ββ (cid:0) s − α − β − s − α (cid:1) ∂ , and consequently (cid:0) s − α − βs − α (cid:1) ∂ + P αj =1 C j (cid:0) s − α − β + js − α (cid:1) ∂ = β (cid:0) s − α − β − s − α − (cid:1) . This means that ( − α + i β r i ( C , . . . , C α ) T is equal to Φ( − β ), withΦ( z ) = ( α − s )Φ ′ ( z ) − Φ ( z ) + ( z + s − α )(Φ ′ ( z ) + Φ ′ ( z ) + Φ ′ ( z ))andΦ ( z ) = P αl,v =0 ( − l + v +1 (cid:0) li (cid:1)(cid:0) vs − α (cid:1)(cid:0) z + s − αv (cid:1)(cid:0) − vl − v (cid:1) , Φ ( z ) = (cid:0) i − s − α − (cid:1)(cid:0) z + i − i − (cid:1) , Φ ( z ) = P αl,j,v =0 ( − α + j + l + v +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) li (cid:1)(cid:0) j − vl − v (cid:1)(cid:0) z + s − α + jv (cid:1)(cid:0) s − α + j − vj − v (cid:1) , Φ ( z ) = (cid:0) α +1 s − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) z + l − l (cid:1) = (cid:0) α +1 s − α (cid:1)(cid:0) z + αα − i (cid:1)(cid:0) z + i − i (cid:1) . So we want to show that Φ( − β ) = 0. If s = α + β then this equation amounts to β Φ ′ ( − β ) + Φ ( − β ) = 0 , and indeed β Φ ′ ( − β ) = β P αl,v =0 ( − l + v +1 (cid:0) li (cid:1)(cid:0) vβ (cid:1)(cid:0) v (cid:1) ∂ (cid:0) − vl − v (cid:1) = P αl,v =0 ( − l βv (cid:0) li (cid:1)(cid:0) vβ (cid:1)(cid:0) − vl − v (cid:1) = P αl,v =0 ( − l (cid:0) li (cid:1)(cid:0) v − β − (cid:1)(cid:0) − vl − v (cid:1) = P αl,v =0 ([ l = 0]( − β + l +1 + [ l = β ]( − l ) (cid:0) li (cid:1) = ( − β (cid:0) βi (cid:1) = [ i = β ]( − β = − (cid:0) i − β − (cid:1)(cid:0) i − β − i − (cid:1) = − Φ ( − β ) . Now suppose that s = α + β . As in the proof of lemma 7 we can simplify Φ ( z )to Φ ( z ) = − (cid:0) z + s − αs − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) z + l − l + α − s (cid:1) . We can also simplify Φ ( z ) toΦ ( z ) = P αj,v =0 ( − α + v +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) vj − i (cid:1)(cid:0) z + s − α + jv (cid:1)(cid:0) s − α + j − vj − v (cid:1) = (cid:0) z + i − i (cid:1) P αj =0 ( − α + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) z + s − α + jj − i (cid:1) . Suppose first that i > β . ThenΦ ′ ( − β ) = − P αl =0 (cid:0) li (cid:1)(cid:16)(cid:0) s − α − βs − α (cid:1) ∂ (cid:0) l − β − l + α − s (cid:1) + (cid:0) s − α − βs − α (cid:1)(cid:0) l − β − l + α − s (cid:1) ∂ (cid:17) , Φ ( − β ) = 0 , Φ ′ ( − β ) = (cid:0) i − β − i (cid:1) ∂ P αj =0 ( − α + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α − β + jj − i (cid:1) , Φ ′ ( − β ) = (cid:0) α +1 s − α (cid:1)(cid:0) α − βα − i (cid:1)(cid:0) i − β − i (cid:1) ∂ . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 13
Thus if s > α + β then the equation Φ( − β ) = 0 is equivalent to L ( s, α, β, i ) = R ( s, α, β, i )with L := P αl =0 (cid:0) lβ +1 (cid:1)(cid:0) l − β − i − β − (cid:1)(cid:0) l − β − s − α − β − (cid:1) ,R := (cid:0) s − αβ +1 (cid:1) P αj =0 ( − α + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α − β + jj − i (cid:1) + (cid:0) α +1 β +1 (cid:1)(cid:0) α − βs − α − β − (cid:1)(cid:0) α − βα − i (cid:1) . Let us in fact show that L ( u, v, w, t ) = R ( u, v, w, t )for all u, v, w, t >
0. We clearly have L ( u, , w, t ) = R ( u, , w, t )since both sides are zero, and R ( u + 1 , v + 1 , w, t ) − R ( u, v, w, t ) − L ( u + 1 , v + 1 , w, t ) + L ( u, v, w, t )= (cid:0) u − vw +1 (cid:1) u − v v − u +2 P vj =0 ( − v + j (cid:0) j v − u +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) u − v − w + jj − t (cid:1) − (cid:0) u − vw +1 (cid:1)(cid:0) v +12 v − u +2 (cid:1)(cid:0) u − w +1 v − t +1 (cid:1) + u − v v − u +2 (cid:0) v +1 w +1 (cid:1)(cid:0) v − wu − v − w − (cid:1)(cid:0) v − w +1 t − w (cid:1) . All we need to show is that this is zero for all u, v, w, t >
0, which follows from P j ( − j (cid:0) j v − u +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) u − v − w + ju − v − w + t (cid:1) = P j,e ( − j (cid:0) v − w +1 u − v − w + i − e (cid:1)(cid:0) j v − u +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) u − v + j − e (cid:1) = P j,e ( − j (cid:0) v − w +1 u − v − w + i − e (cid:1)(cid:0) j v − u + e +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) v − u + e +1 e (cid:1) = P j,e ( − u + e +1 (cid:0) v − w +1 u − v − w + t − e (cid:1)(cid:0) u − v − e − j + u − v − e − (cid:1)(cid:0) v +1 v − j +1 (cid:1)(cid:0) v − u + e +1 e (cid:1) = P e ( − u + e +1 (cid:0) v − w +1 u − v − w + t − e (cid:1)(cid:0) u − v − e − u − v − e (cid:1)(cid:0) v − u + e +1 e (cid:1) = ( − v +1 (cid:0) v − w +1 t − w (cid:1)(cid:0) v +1 u − v (cid:1) . (2)Similarly, if s < α + β then the equation Φ( − β ) = 0 is equivalent to L ( s, α, β, i ) = R ( s, α, β, i )with L := P αl = β +1 (cid:0) ls − α (cid:1)(cid:0) l − β − i − β − (cid:1) ,R := P αj =0 ( − α + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) s − α − β + jj − i (cid:1) + (cid:0) α +1 s − α (cid:1)(cid:0) α − βα − i (cid:1) . Let us in fact show that L ( u, v, w, t ) = R ( u, v, w, t )for all u > v > t > w >
0. It is easy to verify that L ( u, t, w, t ) = R ( u, t, w, t ) , and R ( u + 1 , v + 1 , w, t ) − R ( u, v, w, t ) − L ( u + 1 , v + 1 , w, t ) + L ( u, v, w, t )= u − v v − u +2 P vj =0 ( − v + j (cid:0) j v − u +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) u − v − w + jj − t (cid:1) + u − v v − u +2 (cid:0) v +1 u − v (cid:1)(cid:0) v − w +1 t − w (cid:1) − (cid:0) v +1 u − v − (cid:1)(cid:0) u − w +1 u − v − w + t (cid:1) , which is zero by (2). Finally, suppose that i = β . ThenΦ ′ ( − β ) = − P αl =0 (cid:0) lβ (cid:1)(cid:16)(cid:0) s − α − βs − α (cid:1) ∂ (cid:0) l − β − l + α − s (cid:1) + (cid:0) s − α − βs − α (cid:1)(cid:0) l − β − l + α − s (cid:1) ∂ (cid:17) , Φ ( − β ) = ( − β +1 (cid:0) β − s − α − (cid:1) , Φ ′ ( − β ) = P αj =0 ( − α + β + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1) (cid:16)(cid:0) s − α − β + jj − β (cid:1) ∂ − (cid:0) s − α − β + js − α (cid:1) h β (cid:17) , Φ ′ ( − β ) = ( − β (cid:0) α +1 s − α (cid:1) ( h α − β − h β ) , where h t = 1 + · · · + t is the harmonic number for t ∈ Z > and h t = 0 for t ∈ Z .Since P αj =0 ( − α + β + j +1 (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) z + s − α + js − α (cid:1) = ( − α + β (cid:16)(cid:0) z + s − α − s − α (cid:1) − (cid:0) s − α − s − α (cid:1)(cid:17) , we can simplify Φ ′ ( − β ) toΦ ′ ( − β ) = P αj =0 ( − α (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1)(cid:0) α − s − j − β (cid:1) ∂ − ( − β (cid:16)(cid:0) βs − α (cid:1) − (cid:0) α +1 s − α (cid:1)(cid:17) h β . The equation Φ( − β ) = 0 is therefore equivalent to L ( s, α, β ) = R ( s, α, β )with L := β Φ ′ ( − β ) ,R := ( s − α − β )(Φ ′ ( − β ) + Φ ′ ( − β )) − Φ ( − β ) . Let us show that L ( u, v, w ) = R ( u, v, w ) for all u > v > w >
0. For v = w thisis ( − u w (cid:16)(cid:0) w − u − w (cid:1) ∂ (cid:0) − w − u (cid:1) − (cid:0) w − u − w (cid:1)(cid:0) − w − u (cid:1) ∂ (cid:17) = (2 w − u ) (cid:0) wu − w (cid:1) h w + (cid:0) w − u − w − (cid:1) . If u > w then both sides are zero, if u = 2 w then both sides are 1, and if2 w > u > w then both sides are w ( h w − + w − u ). Thus all we need to do isshow that R ( u + 1 , v + 1 , w ) − R ( u, v, w ) − L ( u + 1 , v + 1 , w ) + L ( u, v, w ) = 0for all u > v > w >
0. By using the equation P j ( − j (cid:0) j v − u +1 (cid:1)(cid:0) v +1 j (cid:1)(cid:0) z + u − v − w + jj − w (cid:1) = ( − v +1 (cid:0) v +1 u − v (cid:1)(cid:0) z + v − w +1 v − w +1 (cid:1) we can get rid of the sum P j and, after some simple algebraic manipulations,simplify this to (cid:0) v +1 w (cid:1) (cid:16)(cid:0) u − v − wu − v (cid:1) ∂ (cid:0) v − w v − u +1 (cid:1) + (cid:0) u − v − wu − v (cid:1)(cid:0) v − w v − u +1 (cid:1) ∂ (cid:17) = ( − w +1 ( u − v − w ) w ( v − w +1) (cid:0) v +1 u − v (cid:1) . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 15
We omit the full tedious details and just mention that since we are able to getrid of the sums P l and P j the aforementioned algebraic manipulations amountto simple cancellations. If u > v + w then (cid:0) v +1 w (cid:1) (cid:16)(cid:0) u − v − wu − v (cid:1) ∂ (cid:0) v − w v − u +1 (cid:1) + (cid:0) u − v − wu − v (cid:1)(cid:0) v − w v − u +1 (cid:1) ∂ (cid:17) = ( − w +1 ( v +1)!( v − w )!( u − v − w )!( w − w !( v − w +1)!(2 v − u +1)!( u − v − w − u − v )! = ( − w +1 ( u − v − w )( v +1)! w ( v − w +1)( u − v )!(2 v − u +1)! = ( − w +1 ( u − v − w ) w ( v − w +1) (cid:0) v +1 u − v (cid:1) , and if u < v + w then (cid:0) v +1 w (cid:1) (cid:16)(cid:0) u − v − wu − v (cid:1) ∂ (cid:0) v − w v − u +1 (cid:1) + (cid:0) u − v − wu − v (cid:1)(cid:0) v − w v − u +1 (cid:1) ∂ (cid:17) = ( − w ( v +1)!( w − v + w − u )!( v − w )! w !( v − w +1)!( u − v )!( v + w − u − v − u +1)! = ( − w +1 ( u − v − w )( v +1)! w ( v − w +1)( u − v )!(2 v − u +1)! = ( − w +1 ( u − v − w ) w ( v − w +1) (cid:0) v +1 u − v (cid:1) . We have finally shown that if i ∈ { β, . . . , α } then r i ( C , . . . , C α ) T = 0 . Computing Θ k,a Throughout the proof we use the results from section 9 of [Ars], which we reproducehere without proofs for convenience.
Lemma 15.
Suppose that α ∈ { , . . . , ν − } .(1) We have ( T − a ) (1 • KZ, Q p θ n x α − n y r − np − α )= P j ( − j (cid:0) nj (cid:1) p j ( p − α ( p ) • KZ, Q p x j ( p − α y r − j ( p − − α − a P j ( − j (cid:0) n − αj (cid:1) • KZ, Q p θ α x j ( p − y r − j ( p − − α ( p +1) + O ( p n ) . (2) The submodule im( T − a ) ⊂ ind GKZ e Σ r contains P i (cid:16)P βl = β − γ C l (cid:0) r − β + li ( p − l (cid:1)(cid:17) • KZ, Q p x i ( p − β y r − i ( p − − β + O (cid:0) ap − β + v C + p p − (cid:1) for all β γ < ν and all families { C l } l ∈ Z of elements of Z p , where v C = min β − γ l β ( v p ( C l ) + l ) . The O (cid:0) ap − β + v C + p p − (cid:1) term is equal to O (cid:0) p p − (cid:1) plus − ap − β p − P βl = β − γ C l p l P = µ ∈ F p [ µ ] − l ( p [ µ ]0 1 ) • KZ, Q p θ n x β − l − n y r − np − β + l . Lemma 16.
Suppose that α ∈ Z and v ∈ Q are such that α ∈ { , . . . , ν − } ,v v p ( ϑ α ( D • )) ,v ′ := min { v p ( a ) − α, v } v p ( ϑ w ( D • )) for α < w < ν − α,v ′ < v p ( ϑ w ( D • )) for w < α. If, for j ∈ Z , ∆ j := ( − j − (1 − p ) − α (cid:0) αj − (cid:1) ϑ α ( D • ) , then v v p ( ϑ α (∆ • )) v p (∆ j ) for all j ∈ Z , and P i (∆ i − D i ) • KZ, Q p x i ( p − α y r − i ( p − − α = [ α s ]( − n +1 D r − sp − • KZ, Q p θ n x r − np − s + α y s − α − n − D • KZ, Q p θ n x α − n y r − np − α + E • KZ, Q p θ α +1 h + F • KZ, Q p h ′ + ERR + ERR , for some ERR and ERR such that ERR ∈ im( T − a ) and ERR = O (cid:0) p ν − v p ( a )+ v + p ν − α (cid:1) , some polynomials h and h ′ , and some E, F ∈ Q p such that v p ( E ) > v ′ and v p ( F ) > v ′ . Lemma 17.
Let { C l } l ∈ Z be any family of elements of Z p . Suppose that α ∈ { , . . . , ν − } and v ∈ Q , and suppose that the constants D i := [ i = 0] C − + [0 < i ( p − < r − α ] P αl =0 C l (cid:0) r − α + li ( p − l (cid:1) satisfy the conditions of lemma 16, i.e. v v p ( ϑ α ( D • )) ,v ′ := min { v p ( a ) − α, v } v p ( ϑ w ( D • )) for α < w < ν − α,v ′ < v p ( ϑ w ( D • )) for w < α. Moreover, suppose that C is a unit. Let ϑ ′ := (1 − p ) − α ϑ α ( D • ) − C − . Suppose that v p ( C − ) > v p ( ϑ ′ ) .(1) If v p ( ϑ ′ ) v ′ then there is some element gen ∈ I a that represents a gen-erator of b N α .(2) If v p ( a ) − α < v then there is some element gen ∈ I a that represents agenerator of a finite-codimensional submodule of T (cid:16) ind GKZ quot( α ) (cid:17) = T (cid:16) b N α / ind GKZ sub( α ) (cid:17) , where T denotes the endomorphism of ind GKZ quot( α ) corresponding to thedouble coset of ( p
00 1 ) . Let us now prove the following additional results.
EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 17
Lemma 18.
Let { C l } l ∈ Z be any family of elements of Z p . Suppose that α ∈ Z and v ∈ Q and the constants D i := [ i = 0] C − + [0 < i ( p − < r − α ] P αl =0 C l (cid:0) r − α + li ( p − l (cid:1) are such that α ∈ { , . . . , ν − } ,v v p ( ϑ α ( D • )) ,v ′ := min { v p ( a ) − α, v } < v p ( ϑ w ( D • )) for w < α. Let ϑ ′ := (1 − p ) − α ϑ α ( D • ) − C − . Then im( T − a ) contains ( ϑ ′ + C − ) • KZ, Q p θ α x p − y r − α ( p +1) − p +1 + C − • KZ, Q p θ n x α − n y r − np − α + P ν − α − ξ = α +1 E ξ • KZ, Q p θ ξ h ξ + F • KZ, Q p h ′ + H, (3) for some h ξ , h ′ , E ξ , F, H such that(1) E ξ = ϑ ξ ( D • ) + O ( p v ) ∪ O ( ϑ α +1 ( D • )) ∪ · · · ∪ O ( ϑ ξ − ( D • )) ,(2) if ξ + α − s ξ − s = 0 then the reduction modulo m of θ ξ h ξ generates N ξ ,(3) v p ( F ) > v ′ , and(4) H = O (cid:0) p ν − v p ( a )+ v + p ν − α (cid:1) and if v p ( a ) − α < v then − pap − α H = g • KZ, Q p θ n x α − n y r − np − α + O (cid:0) p ν − v p ( a ) (cid:1) with g = P λ ∈ F p C ( p [ λ ]0 1 ) + A ( p
00 1 ) + [ r ≡ p − α ] B ( p ) , where A = − C − + P αl =1 C l (cid:0) r − α + ll (cid:1) and B = P αl =0 C l (cid:0) r − α + ls − α (cid:1) . Proof.
This lemma is essentially shown under a stronger hypothesis as lemma 17.The stronger hypothesis consists of the three extra conditions that v p ( ϑ w ( D • )) > min { v p ( a ) − α, v } for all α < w < ν − α , that C ∈ Z × p , and that v p ( C − ) > v p ( ϑ ′ ). These extra con-ditions are not used in the actual construction of the element in (3), rather theyare there to ensure that v p ( E ξ ) > min { v p ( a ) − α, v } for all α < ξ < ν − α , thatthe coefficient of ( p [ λ ]0 1 ) in g is invertible, and that we get an integral element oncewe divide the element( ϑ ′ + C − ) • KZ, Q p θ α x p − y r − α ( p +1) − p +1 + C − • KZ, Q p θ n x α − n y r − np − α by ϑ ′ . Therefore we still get the existence of the element in (3) without these extraconditions, and to complete the proof of lemma 18 we need to verify the propertiesof h ξ , E ξ , F, H, A , and B claimed in (1), (2), (3), and (4). The h ξ and E ξ comefrom the proof of lemma 16, and E ξ • KZ, Q p θ ξ h ξ is X ξ • KZ, Q p P λ =0 [ − λ ] r − α − ξ ( λ ]0 1 )( − θ ) n x r − np − ξ y ξ − n , with the notation for X ξ from the proof of lemma 16. Let E ξ = ( − ξ +1 X ξ . Thencondition (1) is satisfied directly from the definition of X ξ . Let h ξ = ( − ξ +1 P λ =0 [ − λ ] r − α − ξ ( λ ]0 1 )( − θ ) n x r − np − ξ y ξ − n . This reduces modulo m to the element( − ξ P λ =0 [ − λ ] r − α − ξ ( λ ]0 1 ) Y ξ − r = ( − s − α +1 (cid:0) ξ − sξ + α − s (cid:1) X ξ + α − s Y ξ − α of σ ξ − r ( r − ξ ) ∼ = I r − ξ ( ξ ) /σ r − ξ ( ξ ) = quot( ξ ) . This element is non-trivial and generates N ξ if ξ + α − s ξ − s = 0, since then X ξ + α − s Y ξ − α generates N ξ . This verifies condition (2). Condition (3) follows fromthe assumption v ′ < v p ( ϑ w ( D • )) for 0 w < α , as in the proof of lemma 16. Fi-nally, condition (4) follows from the description of the error term in lemma 15, asin the proof of lemma 17. Corollary 19.
Let { C l } l ∈ Z be any family of elements of Z p . Suppose that α ∈ { , . . . , ν − } and v ∈ Q , and suppose that the constants D i := [ i = 0] C − + [0 < i ( p − < r − α ] P αl =0 C l (cid:0) r − α + li ( p − l (cid:1) are such that v v p ( ϑ α ( D • )) ,v ′ := min { v p ( a ) − α, v } v p ( ϑ w ( D • )) for α < w < ν − α,v ′ < v p ( ϑ w ( D • )) for w < α. Suppose also that v p ( a ) Z . Let ϑ ′ := (1 − p ) − α ϑ α ( D • ) − C − , ˇ C := − C − + P αl =1 C l (cid:0) r − α + ll (cid:1) . If ⋆ then ∗ is trivial modulo I a , for each of the following pairs ( ⋆, ∗ ) = ( condition , representation ) . (1) (cid:16) v p ( ϑ ′ ) min { v p ( C − ) , v ′ } , b N α (cid:17) .(2) (cid:16) v = v p ( C − ) < min { v p ( ϑ ′ ) , v p ( a ) − α } , ind GKZ sub( α ) (cid:17) .(3) (cid:16) v p ( a ) − α < v v p ( C − ) & ˇ C ∈ Z × p & C Z × p & α − r > , b N α (cid:17) .(4) (cid:16) v p ( a ) − α < v v p ( C − ) & ˇ C ∈ Z × p , ind GKZ quot( α ) (cid:17) .(5) (cid:16) v p ( a ) − α < v v p ( C − ) & C ∈ Z × p , r (cid:17) , where r is a finite-codimensional submodule of T (ind GKZ quot( α )) . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 19
Proof.
There is one extra condition imposed in addition to the conditions fromlemma 18: that v ′ := min { v p ( a ) − α, v } v p ( ϑ w ( D • )) for α < w < ν − α, and it ensures that v p ( E ξ ) > v ′ for all α < ξ < ν − α . Lemma 18 implies that theelement in (3) is in im( T − a ). Let us call this element γ .(1) The condition v p ( ϑ ′ ) min { v p ( C − ) , v ′ } ensures that if we divide γ by ϑ ′ thenthe resulting element reduces modulo m to a representative of a generator of b N α .(2) The condition v = v p ( C − ) < min { v p ( ϑ ′ ) , v p ( a ) − α } ensures that if we divide γ by C − then the resulting element reduces modulo m to a representative of agenerator of ind GKZ sub( α ).(3, 4, 5) The condition v p ( a ) − α < v v p ( C − ) ensures that the term with thedominant valuation in (3) is H , so we can divide γ by ap − α and obtain the element L + O (cid:0) p ν − v p ( a ) (cid:1) , where L is defined by L := (cid:16)P λ ∈ F p C ( p [ λ ]0 1 ) + A ( p
00 1 ) + [ r ≡ p − α ] B ( p ) (cid:17) • KZ, Q p θ n x α − n y r − np − α with A and B as in lemma 18. This element L is in im( T − a ), and it reducesmodulo m to a representative of (cid:16)P λ ∈ F p C ( p [ λ ]0 1 ) + A ( p
00 1 ) + [ r ≡ p − α ]( − r − α B ( p ) (cid:17) • KZ, F p X α − r . As shown in the proof of lemma 17, if C ∈ Z × p then this element always generatesa finite-codimensional submodule of T (ind GKZ quot( α )) , and if additionally A = 0 (over F p ) then in fact we have the stronger conclusionthat it generates ind GKZ quot( α ) . Suppose on the other hand that C = O ( p ) and A ∈ Z × p . In that case we assumethat 2 α − r > m of L represents a generatorof b N α . 5. Proof of theorem 2
We prove theorem 2 by proving nine propositions which give just enough informa-tion to conclude that Θ k,a is irreducible, but not enough to classify it fully.We assume that r = s + β ( p −
1) + u p t + O (cid:0) p t +1 (cid:1) for some β ∈ { , . . . , p − } and u ∈ Z × p and t ∈ Z > , and we write η = u p t . Asthe main result of [Ars] implies theorem 2 for s > ν , we may assume that s ∈ { , . . . , ν − } . Recall also that we assume ν − < v p ( a ) < ν for some ν ∈ { , . . . , p − } , and that k > p (and consequently r > p ).We now give a list of nine propositions, and show that their union implies theorem 2. Proposition 20. If α < s then ( b N α if β ∈ { , . . . , α − } and α > v p ( a ) − t, ind GKZ sub( α ) otherwiseis trivial modulo I a . Proposition 21. If s α < s and β
6∈ { , . . . , α + 1 } then b N α is trivial modulo I a . Proposition 22. If < α < s then T (ind GKZ quot( α )) if β ∈ { , . . . , α } and α > v p ( a ) − t, b N s − α if β ∈ { , . . . , α } and α < v p ( a ) − t, b N α if β ∈ { α + 1 , . . . , s − α } , b N s − α if β > s − α is trivial modulo I a . Proposition 23. If s α < s and ( α, β ) = ( s , s + 1) then T (ind GKZ quot( α )) if β ∈ { , . . . , s − α } and s − α > v p ( a ) − t,T (ind GKZ quot( α )) if β ∈ { s − α + 1 , . . . , α } and α > v p ( a ) − t, b N α otherwiseis trivial modulo I a . Proposition 24. If α > s then ( T (ind GKZ quot( α )) if α = max { ν − t − , β − } , b N α otherwiseis trivial modulo I a . Proposition 25. If β ∈ { , . . . , s − } and t > ν − s − then b N s/ is trivial modulo I a . Proposition 26. If β ∈ { , . . . , s − } and t = ν − s then b N s/ − is trivial modulo I a . Proposition 27. If β ∈ { s , s + 1 } and t > ν − s − then b N s/ is trivial modulo I a . Proposition 28. If β = s + 1 and t = ν − s − then ind GKZ sub( s + 1) is trivial modulo I a . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 21
Proof that propositions 20–28 imply theorem 2.
Let us assume that Θ k,a is reduciblewith the goal of reaching a contradiction. The classification given by theorem 2in [Ars] implies that Θ k,a has two infinite-dimensional factors, each of which is aquotient of a representation in the set { ind GKZ sub( α ) | α < ν } ∪ { ind GKZ quot( α ) | α < ν } , and moreover that the following classification is true.(1) If the two representations are ind GKZ sub( α ) and ind GKZ sub( α ) then α + α ≡ p − s + 1 . (2) If the two representations are ind GKZ sub( α ) and ind GKZ quot( α ) then α − α ≡ p − . (3) If the two representations are ind GKZ quot( α ) and ind GKZ quot( α ) then α + α ≡ p − s − . The facts that α + α ∈ { , . . . , ν − } ⊆ { , . . . , p − } ,α − α ∈ { − ν, . . . , ν − } ⊆ {− p − , . . . , p − } ,s ∈ { , . . . , ν − } ⊆ { , . . . , p − } imply that the following classification is true as well.(1) If the two representations are ind GKZ sub( α ) and ind GKZ sub( α ) then α + α = s + 1 . (2) If the two representations are ind GKZ sub( α ) and ind GKZ quot( α ) then α = α + 1 . (3) If the two representations are ind GKZ quot( α ) and ind GKZ quot( α ) then α + α = s − . This classification and propositions 20, 21, 22, 23, and 24 together imply that oneof the two representations must be either ind
GKZ sub( s ) or ind GKZ quot( s ), and inthat case the other representation is eitherind GKZ sub( s + 1)(which can only happen if β ∈ { , . . . , s − } and t > ν − s or β ∈ { s , s + 1 } and t > ν − s − GKZ quot( s − s = 2 or β ∈ { , . . . , s − } and t = ν − s ). In the lattercase if s = 2 then either 1 • KZ, Q p x y r − ∈ I a generates ind GKZ quot(0), or ν V k,a is known to be irreducible. Propositions 23, 25, 26, 27, and 28exclude all of the remaining possibilities. Thus if we assume that Θ k,a is reduciblewe reach a contradiction, so Θ k,a must be irreducible. Proof of proposition 20.
First suppose that β > α . We apply part (2) of corol-lary 19 with v = 0 and C j = ( − α (cid:0) s − rα (cid:1) if j = − , j = 0 , ( − α − j (cid:0) s − α +1 α − j (cid:1) if j ∈ { , . . . , α } . Since (cid:0) s − rα (cid:1) = (cid:0) βα (cid:1) + O ( p ) ∈ Z × p , the two conditions we need to verify are v p ( ϑ w ( D • )) > w < α and v p ( ϑ ′ ) > P αj =1 ( − α − j (cid:0) s − α +1 α − j (cid:1) P i> (cid:0) r − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = ( − α ([ w = α ] − [ w = 0]) (cid:0) s − rα (cid:1) + O ( p )(4)for 0 w α . Let F w,j ( z, ψ ) ∈ F p [ z, ψ ] denote the polynomial defined in lemma 11.Since P i> (cid:0) r − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = F w,j ( r, s )by ( c-g ), the conclusion of that lemma when evaluated at z = r and ψ = s im-plies (4). Thus if β > α then we can apply part (2) of corollary 19 and concludethat ind GKZ sub( α ) is trivial modulo I a .Suppose now that β ∈ { , . . . , α − } . If t > v p ( a ) − α then the proof of theorem17 in [Ars] applies here nearly verbatim since (cid:0) s − α +1 α (cid:1) ∈ Z × p , and in fact we can conclude that b N α is trivial modulo I a . So let us suppose that t < v p ( a ) − α . We apply part (2) of corollary 19 with v = t and C j = ( − α (cid:0) s − rα (cid:1) if j = − , j = 0 , ( − α − j (cid:0) s − α +1 α − j (cid:1) + pC ∗ j if j ∈ { , . . . , α } , for some constants C ∗ , . . . , C ∗ α yet to be chosen. Clearly v p ( C − ) = t < v p ( a ) − α, and the other conditions that need to be satisfied in order for corollary 19 to beapplicable are t < v p ( ϑ ′ ) ,t v p ( ϑ w ( D • )) for α w < ν − α,t < v p ( ϑ w ( D • )) for 0 w < α. Let us consider the matrix A = ( A w,j ) w,j α that has integer entries A w,j = P
Let us define C − ( z ) , . . . , C α ( z ) ∈ Z p [ z ] as C j ( z ) = (cid:0) s − z − α +1 (cid:1) if j = − , (cid:0) αs − α − (cid:1) − s − zα +1 if j = 0 , ( − j +1 j +1 (cid:0) s − α − α − j (cid:1) ( z − α ) if j ∈ { , . . . , α } . We apply part (1) of corollary 19 with v = 0 and( C − , C , . . . , C α ) = ( C − ( r ) , C ( r ) , . . . , C α ( r )) . The two conditions we need to verify are v p ( ϑ w ( D • )) > w < α and v p ( ϑ ′ ) = 0.These two conditions follow from the system of equations P αj =0 C j P
First let us assume that β ∈ { , . . . , α } . If we attemptto copy the proof of theorem 17 in [Ars] in this setting, the one place where werun into problems is that some entries of the extended associated matrix N arenot integers (i.e. when we extend the number of rows in A , S , and N to 2 ν − α by defining A w,j , S w,j , and N w,j with the same equations used for the first α + 1rows, we get entries which are not integers). To be more specific, the equation for N w, in this setting is pN w, = (cid:0) s + β ( p − − αw (cid:1) ∂ P i> (cid:0) s + β ( p − − α − wi ( p − − w (cid:1) + O ( p ) , where the second term is O ( p ) because it is still true that P i> (cid:0) r − α − wi ( p − − w (cid:1) − P i> (cid:0) s + β ( p − − α − wi ( p − − w (cid:1) = O ( ǫp ) . On the other hand, P i> (cid:0) s + β ( p − − α − wi ( p − − w (cid:1) = P wl =0 ( − l (cid:0) wl (cid:1) P i> (cid:0) s + β ( p − − α − li ( p − (cid:1) = ( − s − α (cid:0) ws − α (cid:1) + O ( p ) . So A w, = S w, + O ( ǫ ) is integral if w < s − α and A w, = S w, + ( − s − α (cid:0) ws − α (cid:1)(cid:0) s − α − βw (cid:1) ∂ ǫp − + O ( ǫ )if w > s − α . Note that β ∈ { , . . . , α } and s > α by assumption, so S w, is stillalways integral, and if s − α w < ν − α then (cid:0) s − α − βw (cid:1) ∂ = ( − s − α − β − w +1 w ( w − s − α − β ) ∈ Z × p . EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 25
What this means is that if we proceed with the proof of theorem 17 in [Ars] andapply lemma 18 with the constants ( C − , C , . . . , C α ) constructed there such that C is a unit, then we obtain an element( ϑ ′ + C − ) • KZ, Q p θ α x p − y r − α ( p +1) − p +1 + C − • KZ, Q p θ n x α − n y r − np − α + P ν − α − ξ = α +1 E ξ • KZ, Q p θ ξ h ξ + F • KZ, Q p h ′ + H which is in im( T − a ) and is such that v p ( C − ) = v p ( ϑ ′ ) = t + 1 ,v p ( E ξ ) > t + 1 for α + 1 ξ < s − α,v p ( F ) > t + 1 , and with H as in lemma 18. However, v p ( E s − α ) = t and v p ( E ξ ) > t for ξ > s − α .Therefore if t > v p ( a ) − α then the dominant term is H and we can conclude thata submodule of finite codimension in T (ind GKZ quot( α )) is trivial modulo I a , andif t < v p ( a ) − α then the dominant term is E s − α • KZ, Q p θ s − α h s − α and hence b N s − α is trivial modulo I a by part (2) of lemma 18.Now let us assume that β > α . We use the constants constructed in the secondbullet point of the proof of theorem 17 in [Ars], and we apply lemma 18. This givesan element ϑ ′ • KZ, Q p θ α x p − y r − α ( p +1) − p +1 + P ν − α − ξ = α +1 E ξ • KZ, Q p θ ξ h ξ + F • KZ, Q p h ′ + H which is in im( T − a ) and is such that v p ( ϑ ′ ) = 1 ,v p ( E ξ ) > α + 1 ξ < s − α,v p ( F ) > ,v p ( E s − α ) = v p (( r − α ) s − α ) , and with H as in lemma 18. This time the dominant term is either ϑ ′ • KZ, Q p θ α x p − y r − α ( p +1) − p +1 or E s − α • KZ, Q p θ s − α h s − α depending on whether β ∈ { α + 1 , . . . , s − α } or β > s − α . Thus in the former case b N α is trivial modulo I a , and in the latter case b N s − α is trivial modulo I a . Proof of proposition 23.
By proposition 21 we may assume that β
6∈ { , . . . , α + 1 } ,and by proposition 22 we may assume that β = α + 1. If α = s and β ∈ { , . . . , s − α } and s − α < v p ( a ) − t then the claim follows from proposition 22. Thus it is enoughto show that if β ∈ { , . . . , α } then (cid:26) T (ind GKZ quot( α )) if α > v p ( a ) − t, b N α if α < v p ( a ) − t is trivial modulo I a . If α < v p ( a ) − t we apply part (1) of corollary 19, and if α > v p ( a ) − t ) we apply part (5) of corollary 19. In both cases we choose v = t and C j = ( − α + β ( s − α )( α − β +1) β (2 α − s +1) ( αβ ) (cid:0) αs − α (cid:1) ǫ if j = − , j = 0 , ( − j +1 ( s − α − β ) β (cid:0) j α − s +1 (cid:1)(cid:0) α +1 j +1 (cid:1) if j ∈ { , . . . , α } . Since v p ( C − ) = t and C = 1, the conditions we need to verify in order to be ableto apply corollary 19 are t v p ( ϑ w ( D • )) for α w < ν − α,t < v p ( ϑ w ( D • )) for 0 w < α,ϑ ′ = − C − + O ( ǫp ) . Let us consider the matrix A = ( A w,j ) w,j α that has integer entries A w,j = P
1, and( BS ) i,j = (cid:0) s + β ( p − − α + ji ( p − j (cid:1) EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 27 for i ∈ { , . . . , β − } . Let R denote the ( α + 1) × ( α + 1) matrix over F p obtainedfrom BN by replacing the rows indexed 1 , . . . , β − BS . As in the proof of theorem 17 in [Ars] we can compute( BN ) i,j = P αl,v =0 ( − i + l + v (cid:0) li (cid:1)(cid:0) j − vl − v (cid:1)(cid:0) s − α − β + jv (cid:1) ∂ (cid:0) s − α + j − vj − v (cid:1) − [ i = 0] (cid:0) s − α − β + jj (cid:1) ∂ − [ i = β ] (cid:0) s − α − β + js − α (cid:1) ∂ − ( − i (cid:0) s − α − β + js − α (cid:1) P αl =0 (cid:0) li (cid:1)(cid:0) l − β − l (cid:1) ∂ . Thus lemma 14 implies that R ( C , C , . . . , C α ) T = (cid:18) ( − α + β +1 ( s − α )( α − β +1) β (2 α − s +1) ( αβ ) (cid:0) αs − α (cid:1) , , . . . , (cid:19) T . So the conditions we need to apply corollary 19 are indeed satisfied, and thatcompletes the proof.
Proof of proposition 24.
This is the first time that we consider an α such that α > s . The major difference in this scenario is that s is not the “correct” remainderof r to work with and instead we should consider the number that is congruent to r mod p − α + 1 , . . . , p − α −
1. Let us therefore define s α = r − α + α , and in particular let us note that s α = s for s > α (which hashitherto always been the case). Then the computations in the proof of theorem 17in [Ars] work out exactly the same if we replace every instance of s with s α (andthe restricted sum “ P i> ” with “ P ν − α . So there is ananalogous version of theorem 17 in [Ars], and we can conclude the desired result—as the proof of theorem 17 in [Ars] works nearly without modification, we omit thefull details of the arguments. Proof of proposition 25.
Let us write α = s + 1 and, as the claim we want to proveis vacuous for s = 2, let us assume that s > α >
3. We applypart (3) of corollary 19 with v chosen in the open interval ( v p ( a ) − α, t ) and C j = ( j ∈ {− , } , ( − j (cid:0) α − j (cid:1) + ( − j +1 ( α − (cid:0) α − j − (cid:1) + pC ∗ j if j ∈ { , . . . , α } , for some constants C ∗ , . . . , C ∗ α yet to be chosen. The conditions necessary for thelemma to be applicable are satisfied if ˇ C = P j C j (cid:0) r − α + jj (cid:1) ∈ Z × p and ϑ w ( D • ) = O ( ǫ )for 0 w < ν − α . We haveˇ C = P j C j (cid:0) s − α − β + js − α − β (cid:1) + O ( p )= − P j (cid:16) ( − j (cid:0) α − j (cid:1) + ( − j +1 ( α − (cid:0) α − j − (cid:1)(cid:17) (cid:0) s − α − β + js − α − β (cid:1) + O ( p )= − O ( p ) ∈ Z × p by ( c-e ) since α − > s − α − β . And, since j s − α − β + j s − β < p − i, we also have (cid:0) s + β ( p − − α + ji ( p − j (cid:1) = (cid:0) βi (cid:1)(cid:0) s − α − β + jj − i (cid:1) + O ( p ) . Thus the equality ϑ w ( D • ) = O ( p ) follows from the fact that P j ( − j (cid:0) α − j − (cid:1)(cid:0) s − α − β + js − α − β + i (cid:1) = 0 , which follows from ( c-e ) since α − > α − − β + i = s − α − β + i . Moreover, wecan choose C ∗ , . . . , C ∗ α in a way that ϑ w ( D • ) = 0 for 0 w < ν − α similarly as in the proof of theorem17 in [Ars] since the reduction modulo p of the matrix (cid:16)(cid:0) s + β ( p − − α + ji ( p − j (cid:1)(cid:17) i,j<β = (cid:16)(cid:0) βi (cid:1)(cid:0) s − α − β + jj − i (cid:1) + O ( p ) (cid:17) i,j<β is upper triangular with units on the diagonal. Thus the conditions we need toapply corollary 19 are satisfied and we can conclude that b N s/ is trivial modulo I a . Proof of proposition 26.
Let us write α = s − s = 2, let us assume that s > α >
3. Theonly obstruction in the proof of proposition 22 that prevents us from concludingthat b N s/ − is trivial modulo I a is that the dominant terms are E ξ • KZ, Q p θ ξ h ξ for s < ξ ν − s rather than H . We can see from proposition 25 that E s/ • KZ, Q p θ s/ h s/ = x + x , with v p ( x ) > t + 1, and with x ∈ im( T − a ). Since the valuation of the coefficientof H is less than t + 1, we can remove the obstruction coming from E s/ • KZ, Q p θ s/ h s/ by replacing it with x . If s = 2 ν − b N s/ − is trivial modulo I a . Now suppose that s < ν −
2. Thenjust as in the proof of theorem 17 in [Ars] we can apply part (1) of corollary 19 andconclude that b N α is trivial modulo I a as long as ( ǫ, , . . . , T is in the image ofthe matrix A = ( A w,j ) w,j α that has integer entries A w,j = P i> (cid:0) r − α + ji ( p − j (cid:1)(cid:0) i ( p − w (cid:1) = S w,j + ǫN w,j + O ( ǫp )with S and N as in proposition 20. However, this time we can deduce more thanthat: since s < ν − • KZ, Q p x s/ y r − s/ − is equal to g • KZ, Q p θ s/ x s/ − n +1 y r − np − s/ − + x for some g with v p ( g ) > v p ( a ) − s − x ∈ im( T − a ). This in turn byproposition 25 is equal to g • KZ, Q p θ s/ h + x for some g with v p ( g ) > t , some h , and some x ∈ im( T − a ). Here we use thefact that the valuation of the constant C from proposition 25 is at least one andtherefore the corresponding term H is g • KZ, Q p θ s/ x s/ − n +1 y r − np − s/ − + x + O ( ǫ )for some g with v p ( g ) = v p ( a ) − s − x ∈ im( T − a ). In general theerror term would be C ap − s/ g • KZ, Q p θ s/ x s/ − n y r − np − s/ + O ( ǫ )rather than O ( ǫ )—a description of this error term is given in part (2) of lemma 15.This implies that we can add a constant multiple of1 • KZ, Q p x s/ y r − s/ − to the element P i D i • KZ, Q p x i ( p − α y r − i ( p − − α + O ( ap − α )from the proof of lemma 17, and we can translate this back to adding the extracolumn (cid:16)(cid:0) r − αs − α (cid:1) , . . . , (cid:0) rs (cid:1)(cid:17) T to A . As in proposition 20 we can then reduce showing that ( ǫ, , . . . , T is in theimage of A to showing that (1 , , . . . , T is in the image of the ( α + 1) × ( α + 2) matrix R which is obtained from the matrix Q defined in the proof of theorem 17 in [Ars] by replacing all entries in the firstrow with zeros (because this time we do not divide the corresponding row of A by p ) and by adding an extra column corresponding to the extra column of A .Thus, if we index the extra column to be the zeroth column, the lower right α × α submatrix of R is upper triangular with units on the diagonal, the first column of R is identically zero, and all entries of the first row of R except for R , are zero.As when computing ( BN ) i,j in proposition 23 we can find that R , = P αl =0 (cid:0) l − β − l (cid:1) ∂ = Φ ′ ( − β − z ) = P αl =0 (cid:0) z + ll (cid:1) = (cid:0) z + α +1 α (cid:1) . Thus R , = (cid:0) α − βα (cid:1) ∂ = ( − β +1 β ( αβ ) = 0 , which implies that (1 , , . . . , T is in the image of R . Thus the conditions we needto apply corollary 19 are satisfied and we can conclude that b N s/ − is trivial modulo I a . Proof of proposition 27.
Let us write α = s + 1. The reason why the proof ofproposition 25 does not work for β ∈ { α − , α } is because ˇ C = O ( p ) for the con-structed constants C j . However, since t > v p ( a ) − s , if ˇ C ∈ p Z × p then the dominantterm coming from lemma 18 is H = b H ( p
00 1 ) • KZ, Q p θ n x α − n y r − np − α + O (cid:0) p ν − α +1 (cid:1) for the constant b H = ap − α − p ˇ C which has valuation v p ( a ) − α + 1. As in proposition 26 it is crucial here that C = O ( p ). Just as in the proof of proposition 25 we can reduce the claim we wantto show to proving that there exist constants C , . . . , C α ∈ Z p such that C = O ( p )and (cid:16)(cid:0) s + β ( p − − α + ji ( p − j (cid:1)(cid:17) i<β, Let us write α = s + 1. This time the proofs of bothparts (25) and (27) break down since ˇ C = O ( p ) and the dominant term is no longer H . Let us slightly tweak these constants and instead use C j = ( ( − α ǫ if j = − , ( − α + j +1 α (cid:0) α − j − (cid:1) if j ∈ { , . . . , α } . Let R be the matrix constructed in proposition 23. Then just as in the proof ofproposition 25 we can show that ˇ C = O ( p ), and just as in the proof of proposition 20we can show that the dominant term coming from equation (3) in lemma 18 is( ϑ ′ + C − ) • KZ, Q p θ α x p − y r − α ( p +1) − p +1 + C − • KZ, Q p θ n x α − n y r − np − α (and therefore that ind GKZ sub( s + 1) is trivial modulo I a ) as long as R ( C , . . . , C α ) T = (0 , . . . , , T . This follows from lemma 12. Thus the conditions we need to apply corollary 19 aresatisfied and we can conclude that ind GKZ sub( s + 1) is trivial modulo I a . References [Ars] Bodan Arsovski, On the reductions of certain two-dimensional crystalline representa-tions , preprint. https://arxiv.org/abs/1711.03057 .[Ber10] Laurent Berger, Représentations modulaires de GL ( Q p ) et représentations galoisiennesde dimension 2 , Astérisque, Société Mathématique de France (2010), 263–279.[BLZ04] Laurent Berger, Hanfeng Li, and Hui June Zhu, Construction of some families of two-dimensional crystalline representations , Mathematische Annalen (2004), no. 2, 365–377.[BG15] Shalini Bhattacharya and Eknath Ghate, Reductions of Galois representations for slopesin (1 , (2015), 943–987.[Bre03a] Christophe Breuil, Sur quelques représentations modulaires et p -adiques de GL ( Q p ) , I ,Compositio Mathematica (2003), no. 2, 165–188.[Bre03b] , Sur quelques représentations modulaires et p -adiques de GL ( Q p ) , II , Journalof the Institute of Mathematics of Jussieu (2003), no. 1, 23–58.[BG09] Kevin Buzzard and Toby Gee, Explicit reduction modulo p of certain two-dimensionalcrystalline representations , International Mathematics Research Notices (2009),2303–2317.[BG13] , Explicit reduction modulo p of certain two-dimensional crystalline representa-tions, II , Bulletin of the London Mathematical Society (2013), no. 4, 779–788. EDUCTIONS OF CRYSTALLINE REPRESENTATIONS, III 31 [Edi92] Bas Edixhoven, The weight in Serre’s conjectures on modular forms , Inventiones Math-ematicae (1992), 563–594.[GG15] Abhijiit Ganguli and Eknath Ghate, Reductions of Galois representations via the mod p local Langlands correspondence , Journal of Number Theory147