On the boundary components of central streams in the two slopes case
aa r X i v : . [ m a t h . AG ] D ec ON THE BOUNDARY COMPONENTS OFCENTRAL STREAMS IN THE TWO SLOPES CASE
Nobuhiro Higuchi
Abstract
In 2004 Oort studied the foliation on the space of p -divisible groups. In his theory, specialleaves called central streams play an important role. It is still meaningful to investigate centralstreams, for example, there remain a lot of unknown things on the boundaries of central streams.In this paper, we classify the boundary components of the central stream for a Newton polygonconsisting of two segments, where one slope is less than 1 / /
2. Moreover we determine the generic Newton polygon of each boundary component usingthis classification.
In [11, p. 1023], Oort introduced the notion of minimal p -divisible groups, where p -divisible groupsare often called Barsotti-Tate groups. Let k be an algebraically closed field of characteristic p . Oortshowed in [11, 1.2] that they have the following special property: Let X be a minimal p -divisiblegroup over k . Let Y be a p -divisible group over k . If X [ p ] ≃ Y [ p ], then X ≃ Y , where X [ p ] is thekernel of p -multiplication p : X → X . For a Newton polygon ξ , we have a minimal p -divisible group H ( ξ ). See Section 2.1 (2) for the definition of H ( ξ ). A minimal p -divisible group is a p -divisiblegroup which is isomorphic to H ( ξ ) for some ξ over an algebraically closed field.Let X be a p -divisible group over k . Let Def( X ) = Spf(Γ) be the deformation space of X .Here, the deformation space is the formal scheme pro-representing the functor Art k → Set sending R to the set of isomorphism classes of p -divisible groups X over R satisfying X k ∼ = X , where Art k is the category of local Artinian rings with residue field k . It was proved by de Jong in [4, 2.4.4]that the category of p -divisible groups over Spf(Γ) is equivalent to the category of p -divisible groupsover ∆ := Spec(Γ). Let X ′ → Spf(Γ) be the universal p -divisible groups, and let X be the p -divisiblegroups over ∆ obtained from X ′ by the equivalence above.In [10, 2.1], for a p -divisible group Y over k and a k -scheme S, Oort introduced a locally closedsubset C Y (S) for a p -divisible group Y over S characterized by s ∈ C Y (S) if and only if Y s isisomorphic to Y over an algebraically closed field containing k ( s ) and k . He called C Y (S) the leafassociated with Y in S; see (3) in Section 2.1 for the details. We are interested in the case that p -divisible group; 14K10 algebraic moduli, classification. Key words and phrases . p -divisible group; deformation space; Newton polygons. Y , S) = ( X , ∆). In particular, if Y is minimal, he called the leaf the central stream . The notion ofcentral streams is a “central” tool in the theory of foliations.Let X and Y be p -divisible groups over k . We say that X appears as a specialization of Y if there exists a family of p -divisible group X → Spec( R ) with discrete valuation ring ( R, m ) incharacteristic p such that X L is isomorphic to Y over an algebraically closed field containing L and k , and X κ is isomorphic to X over an algebraically closed field containing κ and k , where L = frac R is the field of fractions of R , and κ = R/ m is the residue field of R . We say that a specialization X of Y is generic if ℓ ( X [ p ]) = ℓ ( Y [ p ]) − ℓ ( X [ p ]) is the length of the element of theWeyl group corresponding to X [ p ]; see the paragraph below Corollary 1.2 for this correspondencebetween elements of the Weyl group and p -kernels of p -divisible groups.Let ξ and ζ be Newton polygons. See the paragraph containing (2) for the definition of Newtonpolygons. We say ζ ≺ ξ if each point of ζ is above or on ξ . Moreover, we say that ζ ≺ ξ is saturated if there exists no Newton polygon η such that ζ (cid:22) η (cid:22) ξ .In this paper, we will give two main results. The first result (Theorem 4.1) classifies the boundarycomponents of certain central streams. See Section 4 for the statement as we need some notationgiven in Section 3 to state the first result. From this result, we expect that Conjecture 4.13 statedin Section 4 is true. This conjecture says that it suffices to deal with central streams associated toNewton polygons consisting of two segments in order to classify boundary components of arbitrarycentral streams. The second result is Theorem 1.1.
Let ξ be a Newton polygon consisting of two segments with slopes λ and λ ′ satisfying λ < / < λ ′ . Let X be an arbitrary generic specialization of H ( ξ ). Then there exists a Newtonpolygon ζ satisfying that(i) ζ ≺ ξ is saturated, and(ii) H ( ζ ) appears as a specialization of X .With the same notation as Theorem 1.1, by Grothendieck-Katz [5, 2.3.1], we have ζ ≺ NP( X ) (cid:22) ξ, where NP( X ) is the Newton polygon of X . Here we note that NP( X ) = ξ will be proved inLemma 2.2 below. Then the saturatedness of ζ ≺ ξ implies NP( X ) = ζ and therefore Corollary 1.2.
Let ξ and X be as in Theorem 1.1. Then NP( X ) ≺ ξ is saturated.For a finite flat commutative group scheme G over some base scheme, we say that G is a truncated Barsotti-Tate group of level one (abbreviated as BT ) if Frobenius F and Verschiebung V on G satisfy that Ker F = Im V and Ker V = Im F . The dimension of BT G is defined bylog p rk(Ker F ). Let W be the Weyl group of the general linear group GL h . By Kraft [6], Oort [9]and Moonen-Wedhorn [8], for each algebraically closed field k , there exists a canonical one-to-onecorrespondence { BT ’s over k of height h and dimension d } ↔ J W for a subset J W of W depending on h and d ; see the paragraph below Example 2.6 for the definitionof J W and this correspondence. Let w and w ′ be elements of J W . We say w ′ ⊂ w if there exists adiscrete valuation ring R of characteristic p such that there exists a finite flat commutative groupscheme G → Spec( R ) satisfying that G κ is a BT of type w ′ , and G L is a BT of type w , where L N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 3is the fractional field of R , and κ is the residue field R/ m with the maximal ideal m of R . We saythat w ′ is a generic specialization of w if w ′ ⊂ w and ℓ ( w ′ ) = ℓ ( w ) − p -divisible group X of type w , i.e., for a p -divisible group such that its p -kernel corresponds to w ∈ J W , let S w (∆) bethe reduced subscheme of ∆ = Spec(Γ) consisting of s satisfying that X s [ p ] is of type w . Then it isknown that dim S w (∆) = ℓ ( w ) and S w (∆) is non-empty; see [13, 6.10] and [7, 3.1.6]. This justifiesthe terminology “generic” for elements w of J W .For a Newton polygon ξ , let w ξ be the element of J W corresponding to H ( ξ )[ p ]. Using notationof the Weyl group, by [2, 4.1], Theorem 1.1 is paraphrased as Theorem 1.3.
Let ξ be a Newton polygon consisting of two segments, where one slope is less than1 / /
2. For an arbitrary generic specialization w ∈ J W of w ξ , thereexists a Newton polygon ζ such that(i) ζ ≺ ξ is saturated, and(ii) w ζ ⊂ w .The results of this paper lead us to expect that Theorem 1.3 can be generalized to the case that ξ is an arbitrary Newton polygon: Conjecture 1.4.
For an arbitrary Newton polygon ξ , let w ∈ J W be a generic specialization of w ξ . Then there exists a Newton polygon ζ such that(i) ζ ≺ ξ is saturated, and(ii) w ζ ⊂ w .If Conjecture 4.13 stated in Section 4 holds, then for Newton polygons ξ , the two segments caseis essential to show Conjecture 1.4.This paper is organized as follows: In Section 2, we recall definitions of p -divisible groups,truncated Dieudonn´e modules of level one and related matters. Moreover, we introduce a notionof “arrowed binary sequences”. The proofs of our main results are described using this notion. InSection 3, to show the first result, we give explanations about tools which are used to constructspecializations combinatorially, and show some properties of these tools. In Section 4, we give aproof of Theorem 4.1. This theorem classifies boundary components of central streams. In Section 5,we see the key proposition (Proposition 5.1) which is used to prove the second result, and we showTheorem 1.3. In this section, we recall definitions of p -divisible groups, central leaves, minimal p -divisible groups,central streams and DM ’s. Finally, we introduce the notion of arrowed binary sequences; seeDefinition 2.8, used in the proofs of main theorems. NOBUHIRO HIGUCHI p -divisible groups and Dieudonn´e modules First, let us recall the definition of p -divisible groups. Let p be a prime number. Let h be anon-negative integer. Let S be a scheme in characteristic p . We say that X is a p -divisible group (Barsotti-Tate group) of height h over S if X is an inductive system X = ( G v , i v ) v ≥ for naturalnumbers v , where G v is a finite locally free commutative group scheme over S of order p vh , and foreach v , there exists the exact sequence of commutative group schemes0 → G v i v −→ G v +1 p v −→ G v +1 , where i v is a canonical inclusion. Let X = ( G v , i v ) v ≥ be a p -divisible group over S. For an arbitraryscheme T over S, we have the p -divisible group X T over T which is defined by ( G v × S T , i v × id) v ≥ .In particular, if T is a closed point s over S, then the p -divisible group X s is called fiber of X over s . Let K be a perfect field of characteristic p . We denote by W ( K ) the ring of Witt-vectorswith coefficients in K . Let σ be the Frobenius over K . We denote by the same symbol σ theFrobenius over W ( K ) if no confusion can occur. We say that M is a Dieudonn´e module over K if M is a finite W ( K )-module equipped with σ -linear homomorphism F : M → M and σ − -linearhomomorphism V : M → M satisfying that F ◦ V and V ◦ F is equal to the multiplication by p . Weuse the covariant Dieudonn´e theory, which says that there exists a canonical categorical equivalence D from the category of p -divisible groups (resp. finite commutative group schemes) over K to thatof Dieudonn´e modules over K which are free as W ( K )-modules (resp. are of finite length).Here, let us recall the notion of minimal p -divisible groups. We define a p -divisible group H m,n over F p as follows: H m,n is of dimension n , and its Serre-dual is of dimension m . Moreover, theDieudonn´e module is obtained by(1) D ( H m,n ) = h M i =1 Z p e i , where h = m + n , and Z p is the ring of p -adic integers. For the basis e i , operations F and V satisfythat F e i = e i − m , V e i = e i − n and e i − h = pe i .Let { ( m i , n i ) } i =1 ,...,z be a finite number of pairs of coprime non-negative integers endowed witha non-increasing order of λ i := n i /h i with h i = m i + n i , i.e., we have λ ≥ λ ≥ · · · ≥ λ z . A Newton polygon ξ = P zi =1 ( m i , n i ) is a lower convex polygon in R , breaking on integral coordinates,consisting of slopes λ i . We call each coprime pair ( m i , n i ) segment .For a Newton polygon ξ = P i ( m i , n i ), we set a p -divisible group(2) H ( ξ ) = M i H m i ,n i . We say that a p -divisible group X is minimal if there exists a Newton polygon ξ such that X isisomorphic to H ( ξ ) over an algebraically closed field. For a p -divisible group Y , there exists anisogeny from Y to H ( ξ ) over an algebraically closed field for some Newton polygon ξ . This ξ iscalled the Newton polygon of Y , which is denoted by NP( Y ).In [10, 2.1], for a p -divisible group Y of height h over a scheme S in characteristic p and for a p -divisible group Y over a field of characteristic p , Oort gave the definition of a leaf by(3) C Y (S) = { s ∈ S | Y s is isomorphic to Y over an algebraically closed field } , N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 5which is considered as a locally closed subscheme of S by giving C Y (S) the induced reduced structure.Let K be a field of characteristic p , and let X be a p -divisible group over K. Let Y →
S bea p -divisible group of height h and dimension d over a noetherian scheme S over K. Let ξ be aNewton polygon starting at (0 , h, d ). We write W ξ (S) = { s ∈ S | NP( Y s ) = ξ } . (4)By Grothendieck-Katz [5], we know that W ξ (S) ⊂ S is locally closed. We recall Oort’s result onleaves:
Theorem 2.1 ([10], Theorem 2.2) . Let K be a field, and let X → Spec(K) be a p -divisible groupover K. Set ξ = NP( X ). Let S → Spec(K) be an excellent scheme over K. For a p -divisible group Y → S, C X (S) ⊂ W ξ (S)(5)is a closed subset.Using Theorem 2.1, we show Lemma 2.2.
Let X be a specialization of the minimal p -divisible group H ( ξ ) for a Newton polygon ξ . Assume that X is not isomorphic to H ( ξ ) over an algebraically closed field. Then NP( X ) (cid:22) ξ . Proof.
It suffices to show the case that X is a p -divisible group over an algebraically closed field k of characteristic p . Let R be a discrete valuation ring over k . Let L denote the fractional field of R . Let X → Spec( R ) be a p -divisible group satisfying that X k ≃ X and X L ≃ H ( ξ ) L . Suppose thatNP( X ) = ξ were true. By applying Theorem 2.1 to ( Y →
S) = ( X → Spec( R )) and X = H ( ξ ) k ,we have X ≃ H ( ξ ) k . This is a contradiction.For a p -divisible group X , the kernel of p -multiplication p : X → X is called the p -kernel of X , denoted by X [ p ]. The Dieudonn´e module of X [ p ] makes a truncated Dieudonn´e module of levelone, defined below: Definition 2.3.
Let N be a K -vector space of finite dimension. Let F and V be a σ -linear mapand a σ − -linear map respectively from N to itself. A triple ( N, F , V) is a truncated Dieudonn´emodules of level one over K (abbreviated as DM ) if the above F and V satisfy that Ker F = Im Vand Im F = Ker V. We say that a DM ( N, F , V) is of height h and dimension d if dim K N = h anddim K N/ V N = d .Let us recall the notion of specializations. Let R be a commutative ring of characteristic p > σ : R → R be the Frobenius endomorphism defined by σ ( a ) = a p . Then the definition of DM ’sover the ring R is given as follows: Definition 2.4.
A DM over R of height h is a quintuple N = ( N , C, D, F, V − ) satisfying(i) N is a locally free R -module of rank h ,(ii) C and D are submodules of N which are locally direct summands of N ,(iii) F : ( N /C ) ⊗ R,σ R → D and V − : C ⊗ R,σ R → N /D are R -linear isomorphisms. NOBUHIRO HIGUCHIPut R = K [[ t ]]. Let N be an arbitrary DM over R . We set N K = N ⊗ R K , and we have a DM over K . From this we obtain a map, called a specialization , { DM over R } → { DM over K } which maps N to N K .Let ξ = P i ( m i , n i ) be a Newton polygon. We denote by N ξ the DM associated to the p -kernelof H ( ξ ). Equivalently, N ξ is described as(6) N ξ = M N m i ,n i , where N m,n is the DM associated to the p -kernel of H m,n . In this paper, we mainly treat DM ’s N ξ with ξ = ( m , n ) + ( m , n ) satisfying λ < / < λ .Let k be an algebraically closed field of characteristic p . Then the following classification ofDM ’s over k is given by Kraft [6], Oort [9] and Moonen-Wedhorn [8]. Theorem 2.5.
There exists a bijection: { , } h ↔ { DM over k of height h } / ∼ = . Let us review the construction of the bijection above. We often identify an element S of { , } h with the pair of a totally ordered set ˜ S = { t < · · · < t h } and a map δ : ˜ S → { , } , so that the i -th coordinate of S is δ ( t i ). We write this identification as an equality S = ( ˜ S, δ ). The bijectionof Theorem 2.5 is obtained in the following way. Let S = ( ˜ S, δ ) be as above. To S , we associate aDM ( N, F , V) as follows. Let N = ke ⊕ · · · ⊕ ke h . We define a map F byF e i = ( e j , j = { l | δ ( t l ) = 0 , l ≤ i } for δ ( t i ) = 0 , . Let t j , . . . , t j c , with j < · · · < j c , be the elements of ˜ S satisfying δ ( t j l ) = 1. Put d = h − c . Thena map V is defined by V e i = ( e j l , l = i − d for i > d, . We call { e , . . . , e h } a standard basis of ( N, F , V).
Example 2.6.
Let us see an example of DM ’s. Let S = (1 , , , ,
0) be an element of { , } .Then the DM ( N, F , V), with N = ke ⊕ · · · ⊕ ke , corresponding to S is given by the followingdiagram. 1 1 0 F } } V a a V a a F y y V X X e e e e e For the above diagram, if there exists no vector of F (resp. V) from e i , then we regard F (resp. V)maps e i to zero. One can check that the above satisfies the condition of DM ’s.N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 7Let h and c be non-negative integers. Put d = h − c . Let W be the Weyl group of the generallinear group GL h . We identify W with the symmetric group S h . Here, we associate DM ’s over k with elements of the Weyl group. Let s i ∈ W be the simple reflection ( i, i + 1) for i = 1 , . . . , h − S h , namely Ω = { s , . . . , s h − } . Set J c = Ω \ { s c } . Let W J bethe subgroup of W generated by J = J c . Let J W be the set consisting of elements of minimal lengthin W J \ W , i.e., the shortest representatives of W J \ W . Let w be an element of J W . We associate S = ( ˜ S, δ ) to w by the pair of a totally ordered set ˜ S = { t < · · · < t h } and the map δ : ˜ S → { , } defined by δ ( t i ) = 0 if and only if w ( i ) > c . Here the property of the minimal length of w is used.We regard this pair ( ˜ S, δ ) as the element of { , } h . One can check that this gives a one-to-onecorrespondence between J W and the set of elements S of { , } h satisfying { t ∈ ˜ S | δ ( t ) = 0 } = d .Thus there exists a bijection between J W and the set of isomorphism classes of DM ’s over k ofheight h and dimension d . We denote by w ξ the element of J W associated to N ξ .In the rest of this subsection, we show a lemma used for the construction of generic special-izations. Let W and J = J c be as above. Set d = h − c . We define x ∈ W to be x ( i ) = i + d if i ≤ c and x ( i ) = i − c otherwise. We define θ : W → W by u xux − . It follows from [12,4.10] by Viehmann-Wedhorn that w ′ ⊂ w if and only if there exists an element u of W J such that u − w ′ θ ( u ) ≤ w , where ≤ denotes the Bruhat order. Lemma 2.7.
Let w ′ and w be elements of J W with w ′ ⊂ w . If ℓ ( w ′ ) = ℓ ( w ) −
1, then there exist v ∈ W and u ∈ W J such that(i) v = ws for a transposition s ,(ii) ℓ ( v ) = ℓ ( w ) − w ′ = uvθ ( u − ). Proof.
Let w ∈ J W . Let w ′ ∈ J W satisfying that w ′ ⊂ w and ℓ ( w ′ ) = ℓ ( w ) −
1. Choose anelement u of W J satisfying that u − w ′ θ ( u ) < w . Set v = u − w ′ θ ( u ). We show (ii). Since w ′ is an element of J W , we have ℓ ( v ) ≥ ℓ ( u − w ′ ) − ℓ ( θ ( u ) − ) = ℓ ( u ) + ℓ ( w ′ ) − ℓ ( θ ( u )). Moreover, ℓ ( u ) + ℓ ( w ′ ) − ℓ ( θ ( u )) = ℓ ( w ′ ) since for all element u ′ of W J we have ℓ ( u ′ ) = ℓ ( θ ( u ′ )) by thedefinition of θ . As v < w , we have ℓ ( v ) < ℓ ( w ). These prove (ii). Let w = s i s i . . . s i l be a reducedexpression of w with v = s i . . . s i q − s i q +1 . . . s i l . Set s = ( s i l . . . s i q +1 ) s i q ( s i q +1 . . . s i l ). Then s is atransposition, and this s satisfies v = ws . To show our main results, we introduce arrowed binary sequences. This object can be regarded asa generalization of DM ’s. Definition 2.8. An arrowed binary sequence (we often abbreviate as ABS) is the triple ( ˜ S, δ, π )consisting of a totally ordered set ˜ S = { t < · · · < t h } , a map δ : ˜ S → { , } and a bijection π : ˜ S → ˜ S . We denote by H the set of all arrowed binary sequences. Definition 2.9.
Let N = ( N, F , V) be a DM of height h . Let e , . . . , e h be a standard basis of N . Let S = ( ˜ S, δ ) be the element of { , } h corresponding to N with ˜ S = { t , . . . , t h } . We define abijective map π : ˜ S → ˜ S by π ( t i ) = t j , where j is uniquely determined by(7) ( F( e i ) = e j if δ ( t i ) = 0 , V( e j ) = e i otherwise . NOBUHIRO HIGUCHINote that ( ˜
S, δ, π ) is an ABS. We say that an ABS S = ( ˜ S, δ, π ) is admissible if there exists a DM N = ( N, F , V) such that ( ˜
S, δ ) corresponds to N by Theorem 2.5, and π is constructed by (7) from N . The admissible ABS S obtained from a DM N is called the ABS associated to N . We denoteby H ′ the set of all admissible ABS’s.For an arrowed binary sequence S = ( ˜ S, δ, π ), as seen in Example 2.10, we obtain a diagram ofthe ABS using elements of ˜ S and arrows • • , π (cid:2) (cid:2) • π C C • . Example 2.10.
The diagram of the ABS corresponding to (1 , , , ,
0) is1 π > > π > > π ~ ~ π G G π z z . From Example 2.6 and this diagram, one can check that the admissible ABS ( ˜
S, δ, π ) is obtainedby a DM .Now we show a property of admissible ABS’s. Let S = ( ˜ S, δ, π ) be an ABS. For each element t of ˜ S , we define the binary expansion b ( t ) by b ( t ) = 0 .b b · · · , where b i = 0 if δ ( π − i ( t )) = 0, and b i = 1 otherwise. Lemma 2.11.
Let S = ( ˜ S, δ, π ) be an admissible ABS. Let t and t ′ be elements of ˜ S . Then thefollowing holds.(i) Suppose δ ( t ) = δ ( t ′ ). Then t < t ′ if and only if π ( t ) < π ( t ′ ),(ii) Suppose b ( t ) = b ( t ′ ). Then t < t ′ if and only if b ( t ) < b ( t ′ ). Proof. (i) follows from the construction of π defined in Definition 2.9.Let us see (ii). To show “only if” part, we suppose b ( t ) > b ( t ′ ). It implies that there exists anatural number v such that δ ( π − u ( t )) = δ ( π − u ( t ′ )) for all u with u < v and δ ( π − v ( t )) > δ ( π − v ( t ′ )).Then we clearly have δ ( π − v ( t )) = 1 and δ ( π − v ( t ′ )) = 0. By the construction of π , we have π − v +1 ( t ) > π − v +1 ( t ′ ), and this contradicts with (i). Let us see “if” part. If b ( t ) < b ( t ′ ), then thereexists a natural number v such that δ ( π − u ( t )) = δ ( π − u ( t ′ )) for u < v and δ ( π − v ( t )) < δ ( π − v ( t ′ )).Then we have δ ( π − v ( t )) = 0 and δ ( π − v ( t ′ )) = 1, and π − v +1 ( t ) < π − v +1 ( t ′ ) holds. Using (i)repeatedly, we see t < t ′ .Let us define special DM ’s which correspond to special p -divisible groups H m,n . Definition 2.12.
We say a DM N is DM -simple if there exist coprime natural numbers h and m such that N is associated to the ABS ( { t , . . . , t h } , δ, π ), where δ ( t i ) = 1 if i ≤ m and δ ( t i ) = 0otherwise, with the map π defined as Definition 2.9. In other words, we have π ( t i ) = t i − m mod h .That kind of DM corresponds to N m,n with n = h − m , and we call this DM simple DM if noconfusion can occur.N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 9Here, let us recall the direct sum of admissible ABS’s, which corresponds to the direct sum ofassociated Dieudonn´e modules. We define the direct sum S = ( ˜ S, δ, π ) of elements A = ( ˜ A, δ A , π A )and B = ( ˜ B, δ B , π B ) of H ′ as follows. We define a set ˜ S by ˜ S = ˜ A ⊔ ˜ B . We define a map δ : ˜ S → { , } to be δ | ˜ A = δ A and δ | ˜ B = δ B , and let π : ˜ S → ˜ S be a map satisfying that π | ˜ A = π A and π | ˜ B = π B . We define an order < in ˜ S so that(i) for elements t, t ′ ∈ ˜ S , if b ( t ) ≤ b ( t ′ ), then t < t ′ , and(ii) there is no elements t, t ′ of ˜ S such that t < t ′ and π ( t ′ ) < π ( t ),where b ( t ) is the binary expansion of t ∈ ˜ S determined by π . For instance, if A = B with ˜ A = { t , . . . , t h } and ˜ B = { t ′ , . . . , t ′ h } , then ˜ S = { t , t ′ , . . . , t h , t ′ h } . Thus we get the ABS A ⊕ B = ( ˜ S, δ, π )which also belongs to H ′ . Let M and N be DM ’s, and let A (resp. B ) be the ABS correspondingto M (resp. N ). Then the ABS A ⊕ B corresponds to the direct sum of DM ’s M ⊕ N . In thispaper, we consider the case that M = N m ,n and N = N m ,n , with pairs of coprime non-negativeintegers ( m , n ) and ( m , n ). Example 2.13.
Let M = N , and N = N , . Let A and B be ABS’s corresponding to M and N respectively. Then diagrams of these ABS’s are given by the following: A = 1 (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) , (cid:4) (cid:4) B = 1 @ @ @ @ @ @ @ @ @ @ z z z z z z . We write t for the element of ˜ A corresponding to the underlined element in the diagram. Then thebinary expansion b ( t ) is given by b ( t ) = 0 . · · · . In the same way, we obtain the binaryexpansions of all elements of ˜ A and ˜ B . We sort all elements by the binary expansions, and thedirect sum A ⊕ B is given by the following:1 A A A (cid:4) (cid:4) A (cid:4) (cid:4) A (cid:4) (cid:4) A (cid:4) (cid:4) A (cid:4) (cid:4) B B B A y y A y y B > > B > > B w w B w w B w w , where τ Ai (resp. τ Bi ), with τ = 0 or 1, is the i -th element of A (resp. B ).For certain DM ’s, we have the following: Lemma 2.14.
Let ξ = ( m , n ) + ( m , n ) be a Newton polygon satisfying λ < / < λ . Set h = m + n and h = m + n . Let N ξ be the minimal DM of ξ . For the above notation, theABS S associated to N ξ is obtained by the following:1 A · · · Am | {z } m Am +1 · · · An | {z } n − m B · · · Bn | {z } n An +1 · · · Ah | {z } m Bn +1 · · · Bm | {z } m − n Bm +1 · · · Bh | {z } n . Proof.
See [1], Proposition 4.20.0 NOBUHIRO HIGUCHI
In this section, we give a method to construct a specialization of a DM using arrowed binarysequences. We introduce some sets which help us to investigate properties of the specializationobtained by this method. These properties are useful for classification of boundary components ofcentral streams, given in Section 4. Here, we prepare some notation for describing a specialization of a DM in arrowed binary sequences.First, we define the lengths of ABS’s. We use some notation of Section 2.2. Definition 3.1.
Let S = ( ˜ S, δ, π ) be an ABS. We define the length of S by ℓ ( S ) = { ( t ′ , t ) | δ ( t ′ ) = 0 and δ ( t ) = 1 with t ′ < t } . Remark 3.2.
If an ABS S corresponds to a DM N , then the value ℓ ( S ) is equal to the length ℓ ( w ) for the element w of J W , which corresponds to N . In other words, the length of w can becalculated by the above ℓ ( S ). Example 3.3.
For the ABS A ⊕ B associated to N , ⊕ N , , which is constructed in Example 2.13,we have ℓ ( A ⊕ B ) = 29. In general, for a Newton polygon ξ = ( m , n ) + ( m , n ) of two segmentswith λ < / < λ , the length ℓ ( S ) of the ABS S corresponding to N ξ is equal to m n − m n .Let H ′ ( h, d ) denote the set of admissible ABS’s whose corresponding DM ’s are of height h anddimension d . By translating the ordering ⊂ on J W via the bijection from J W to H ′ ( h, d ), we havean ordering on H ′ ( h, d ) as well as the notion of specialization of admissible ABS’s.Here we give a method to construct a type of the specializations of ABS’s. Those will turn outto correspond to specializations of the form w ′ ⊂ w with v = ws < w and w ′ = uvθ ( u − ), where s isa transposition and u ∈ W J . See the paragraph before Lemma 2.7 for this specialization. Startingwith S = ( ˜ S, δ, π ) ∈ H ′ ( h, d ), we construct a new admissible ABS S ′ = ( ˜ S ′ , δ, π ). First we considera “small modification” of π : Definition 3.4.
Let S = ( ˜ S, δ, π ) be an ABS with ˜ S = { t < · · · < t h } . Choose elements t i and t j of ˜ S . We define a map π on ˜ S by π ( π − ( t i )) = t j , π ( π − ( t j )) = t i and π ( t ) = π ( t ) for the otherelements t of ˜ S . We call this map π a small modification by t i and t j of π .We require δ ( t i ) = 0, δ ( t j ) = 1 and t i < t j , when we consider specializations. Let π : ˜ S → ˜ S bethe small modification by elements t i and t j of π . Let ˜ S ′ = ˜ S as sets. There uniquely exists anordering < ′ on ˜ S ′ so that(i) If t < ′ t ′ , then b ( t ) ≤ b ( t ′ ).(ii) Suppose δ ( t ) = δ ( t ′ ). Then t < ′ t ′ if and only if π ( t ) < ′ π ( t ′ ),where b ( t ) denotes the binary expansion determined by δ and π . Then we get an admissible ABS S ′ = ( ˜ S ′ , δ, π ); see Remark 3.5. We call this S ′ the specialization of S obtained by exchanging t i and t j . We say an exchange of t i and t j is good if the specialization S ′ obtained by exchanging t i and t j satisfies ℓ ( S ′ ) = ℓ ( S ) −
1. We often write S − for S ′ when S ′ is the specialization obtained by agood exchange. We call S − a generic specialization of S . Any good exchange produces a genericspecialization, and conversely every generic specialization is obtained by a good exchange. Thisfollows from Lemma 2.7 and Remark 3.5 below:N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 11 Remark 3.5.
We use the same notation as Section 2.1. Let S = ( ˜ S, δ, π ) ∈ H ′ ( h, d ) with ˜ S = { t < · · · < t h } . Let S ′ = ( ˜ S ′ , δ, π ) be the specialization of S obtained by exchanging t i and t j . Let w be the element of J W corresponding to S . Set s = ( i, j ) the transposition. We regard π and π as elements of W . Then π = xw . Put ˜ S ′ = { t ′ < ′ · · · < ′ t ′ h } . Set ˜ S (0) = { t (0)1 < ′′ · · · < ′′ t (0) h } tobe t s ( z ) = t (0) z . We define an element ε of W to be t (0) ε ( z ) = t ′ z . Note that ε stabilizes { , , . . . , d } (and therefore { d + 1 , . . . , h } ), since b ( t (0) z ) < . z ≤ d and b ( t (0) z ) > . z > d , where binaryexpansions are determined by δ and π . Put v = ws . Then w ′ = uvθ ( u − ) corresponds to S ′ with u = x − ε − x ∈ W J . The map π is obtained by ε − π sε . In terminologies of ABS’s, multiplying π by s corresponds to constructing the small modification. By the coordinate transformation by ε ,we obtain the map π on ˜ S ′ . The main purpose of Section 3.2 is to give the construction of this ε combinatorially. Example 3.6.
Let us see an example of constructing a specialization of an ABS. Let S = ( ˜ S, δ, π )be the admissible ABS associated to (1 , , , , , , , S is described as the following dia-gram. 1 (cid:9) (cid:9) (cid:9) (cid:9) > > (cid:4) (cid:4) D D ~ ~ ~ ~ . Let π be the small modification map by 0 and 1 . Then the ABS S ′ = ( ˜ S ′ , δ, π ) is described as1 (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) I I ~ ~ . One can check that this specialization is generic by calculating lengths of S and S ′ .Let ξ = ( m , n )+( m , n ) be a Newton polygon satisfying that λ < / < λ . Let S denote theABS of N ξ . To classify generic specializations of S , in Section 3.2 we introduce a method to constructspecializations S ′ combinatorially. The method gives an explicit construction of permutations ε asin Remark 3.5. Moreover, we show some properties of generic specializations; see Section 4.2. Theseproperties are useful for to show Proposition 5.1 which is a key step of the proof of Theorem 1.3.Let A and B be ABS’s corresponding to N m ,n and N m ,n respectively. Then S is obtained by A ⊕ B . We often denote by τ Ai (resp. τ Bi ), with τ = 0 or 1, the i -th symbol of A (resp. B ). ByLemma 2.14, there exist only three cases as the choice of exchange of 0 Ai and 1 Bj :(1) m < i ≤ n and 1 ≤ j ≤ n ;(2) m < i ≤ n and n < j ≤ m ;(3) n < i ≤ h and n < j ≤ m .We assume that m + 1 < n or n + 1 < m in the case (2). For ξ = ( m , m + 1) + ( n + 1 , n )with m > n >
0, the case (2) is treated separately; see below the proof of Proposition 3.20.Since the case (3) can be regarded as the dual of (1), it suffices to deal with the case (1) and (2).From now on, we assume m < i ≤ n . We often treat the cases (1) and (2) simultaneously, but forexample, the proof of Proposition 3.20 is divided into the cases (1) and (2).2 NOBUHIRO HIGUCHI Let ξ = ( m , n ) + ( m , n ) be a Newton polygon of two segments satisfying λ < / < λ .Let S = ( ˜ S, δ, π ) be the ABS associated to the minimal DM N ξ . In this section, we introduce acombinatorial method to construct a specialization S ′ of S in order to classify generic specializations.Concretely, for the small modification π by t i and t j , we shall construct the ordered set ˜ S ′ , whichcoincides with ˜ S as sets, satisfying that t < t ′ if and only if π ( t ) < π ( t ′ ) for elements t and t ′ of ˜ S ′ with δ ( t ) = δ ( t ′ ). For this ordered set, using Theorem 2.5, we get the specialization N ′ ξ of N ξ . Themethod gives a useful characterization to ABS’s S ′ satisfying ℓ ( S ′ ) = ℓ ( S ) −
1. The main purposeof this section is to show that this combinatorial operation is well-defined.Let A (resp. B ) be the ABS corresponding to N m ,n (resp. N m ,n ). Then S = A ⊕ B . Wedenote by τ Ai (resp. τ Bj ) the i -th element of A (resp. the j -th element of B ) with τ = 0 or 1.Fix natural numbers i and j satisfying that m < i ≤ n and 1 ≤ j ≤ m . Let π be the smallmodification by 0 Ai and 1 Bj . For the ordered set˜ S = { A < · · · < Ai − < Ai < Ai +1 < · · · < Bj − < Bj < Bj +1 < · · · < Bh } , we define an ordered set S (0) by S (0) = ˜ S as sets with the ordering S (0) = { A < · · · < Ai − < Bj < Ai +1 < · · · < Bj − < Ai < Bj +1 < · · · < Bh } . In Definition 3.7 and Definition 3.8, we construct a sequence ( S (0) , δ, π ) , ( S (1) , δ, π ) , . . . of ABS’s.We will show that there exists a non-negative integer n such that the ABS ( S ( n ) , δ, π ) coincides withthe specialization of S obtained by exchanging 0 Ai and 1 Bj . Definition 3.7.
For the ABS ( S (0) , δ, π ), we define a set A (0) by A (0) = { t ∈ S (0) A | t < Ai and π (0 Ai ) < π ( t ) in S (0) with δ ( t ) = 0 } , where S (0) A is a subset of S (0) consisting of all elements of A . For non-negative integers n , put α n = π n (0 Ai ). For a natural number n , we construct an ordered set S ( n ) and a set A ( n ) from theABS ( S ( n − , δ, π ) and the set A ( n − if A ( n − is not empty. Let S ( n ) = S ( n − as sets. We definethe order on S ( n ) to be for t < t ′ in S ( n − , we have t > t ′ in S ( n ) if and only if α n < t ′ ≤ π ( t max )in S ( n − and t = α n . We can regard π as the map on S ( n ) . Thus we obtain an ABS ( S ( n ) , δ, π ).In other words, we define the ABS ( S ( n ) , δ, π ) by moving the element α n to the right of π ( t max ) in S ( n − . We define A ( n ) = { t ∈ S ( n ) A | t < α n and α n +1 < π ( t ) in S ( n ) with δ ( t ) = δ ( α n ) } , where S ( n ) A is the subset of S ( n ) consisting of all elements of A .We will see that there exists a non-negative integer a satisfying A ( a ) = ∅ in Proposition 3.14.For this number a , we introduce a definition of sets B ( n ) as follows. Definition 3.8.
We define the set B (0) = { t ∈ S ( a ) | Bj < t and π ( t ) < π (1 Bj ) in S ( a ) with δ ( t ) = 1 } , and let I be the subset of B (0) consisting of elements t which are of the form t = 1 Ax with a naturalnumber x . Note that for an element t of I , there exists a non-negative integer n with n < a suchN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 13that t = α n . We have 1 Bj < Bz and π (1 Bz ) < π (1 Bj ) for natural numbers z with z < j in S ( a ) . Thusthe set B (0) is determined as I ∪ { B , . . . , Bj − } . Set β n = π n (1 Bj ) for non-negative integers n . Weconstruct the ordered set S ( a + n ) for natural numbers n inductively. For the ABS ( S ( a + n − , δ, π )and the set B ( n − , if B ( n − is not empty, then let S ( a + n ) = S ( a + n − as sets. The ordering of S ( a + n ) is given so that for t < t ′ in S ( a + n − , we have t > t ′ in S ( a + n ) if and only if π ( t min ) ≤ t < β n in S ( a + n − and t ′ = β n . We can regard π as the map on S ( a + n ) . Thus we obtain an ABS ( S ( a + n ) , δ, π ).In other words, we obtain the ABS ( S ( a + n ) , δ, π ) by moving the element β n to the left of π ( t min ) in S ( a + n − . We define B ( n ) = { t ∈ S ( a + n ) | β n < t and π ( t ) < β n +1 in S ( a + n ) with δ ( t ) = δ ( β n ) } . If there exists a non-negative integer b such that B ( b ) = ∅ , then we call the ABS ( S ( a + b ) , δ, π )the full modification by Ai and Bj . If there exists the full modification by 0 Ai and 1 Bj , then thespecialization of S obtained by exchanging 0 Ai and 1 Bj is given by this full modification. We showthat there exists a non-negative integer b such that B ( b ) = ∅ in Proposition 3.20. Example 3.9.
Let us see an example of constructing S ′ from S . Let N = N , ⊕ N , be the DM ,and let S = ( ˜ S, δ, π ) denote the ABS associated to N . In Example 2.13, we obtain the diagram of S . Construct the small modification by 0 A and 1 B . Then we obtain the ABS ( S (0) , δ, π ), and thediagram of this ABS is described as( S (0) , δ, π ) : 1 A A A (cid:4) (cid:4) A (cid:4) (cid:4) A (cid:4) (cid:4) B ◦ A × (cid:4) (cid:4) B × : : B × : : A ◦ } } A z z A z z B ? ? B ? ? B w w B w w B w w . First, let us construct sets A ( n ) . We have the set A (0) = { A } . To construct the ordered set S (1) ,we move the element 0 A to the right of 0 A , and we obtain the following diagram of ( S (1) , δ, π ):( S (1) , δ, π ) : 1 A A A (cid:4) (cid:4) A × (cid:10) (cid:10) A ◦ (cid:6) (cid:6) B ◦ A (cid:127) (cid:127) B × : : B × : : A z z A z z A z z B ? ? B ? ? B w w B w w B w w . We have the set A (1) = { A } . In the same way, we get S (2) by moving the element 1 A to the rightof 0 A , and we see that A (2) is empty. Hence we obtain a = 2. Next, let us construct sets B ( n ) . Wehave B (0) = { B , B } . We move the element 0 B to the left of 1 B which is the minimum element of π ( B (0) ) to construct S (3) . Then B (1) = ∅ . Hence we get an ABS S ′ by ( S (3) , δ, π ). Thus the diagramof the specialization S ′ is described as follows. S ′ : 1 A A (cid:10) (cid:10) A A (cid:4) (cid:4) A (cid:4) (cid:4) B A (cid:127) (cid:127) B B A z z A z z A z z B z z B D D B D D B w w B w w . One can check that these admissible ABS’s S ′ and S satisfy that ℓ ( S ′ ) = ℓ ( S ) −
1. Moreover, one can4 NOBUHIRO HIGUCHIsee that for the same notation as Remark 3.5, the permutation ε satisfies ε − = (13 , , , , S , we show some properties of ordered sets S ( n ) and sets A ( n ) in Lemma 3.10, Proposition 3.11, Corollary 3.12 and Proposition 3.13. For a Newton polygon ξ =( m , n ) + ( m , n ) consisting of two segments satisfying λ < / < λ , we denote by S = ( ˜ S, δ, π )the ABS associated to the DM N ξ . Let A (resp. B ) be the ABS corresponding to the DM N m ,n (resp. N m ,n ). We denote by ˜ A (resp. ˜ B ) the ordered set of A (resp. B ). Let π denote thesmall modification by elements 0 Ai and 1 Bj of ˜ S with m < i ≤ n . Then we obtain sets A ( n ) fornon-negative integers n . We will show that there exists a non-negative integer n such that A ( n ) = ∅ in Proposition 3.14. Lemma 3.10.
For every non-negative integer n , let Σ ( n ) be the set consisting of elements t of ˜ A satisfying that t ′ < t and π ( t ) < π ( t ′ ) in S ( n ) for some elements t ′ of S ( n ) with δ ( t ) = δ ( t ′ ). If A ( n ) is not empty, then Σ ( n ) = { α n } , where α n = π n (0 Ai ). Moreover, if A ( n ) = ∅ , then Σ ( n ) = ∅ . Proof.
Let us show the assertion by induction on n . The case n = 0 is obvious. For a naturalnumber n , suppose that Σ ( n − = { α n − } . The statement follows from the construction of the set S ( n ) . In fact, by the construction of S ( n ) , we have π ( t max ) < α n in S ( n ) , where t max is the maximumelement of A ( n − . It implies that π ( t ) < α n holds for all elements t of A ( n − in S ( n ) . Then α n − does not belong to Σ ( n ) . Thus we see that elements t ′′ of ˜ A \ { α n } do not belong to Σ ( n ) , and weobtain Σ ( n ) = { α n } . Proposition 3.11.
For all non-negative integers m , set α m = π m (0 Ai ). Let n be a natural number.Then A ( n ) is obtained by A ( n ) = { π ( t ) ∈ ˜ A | t ∈ A ( n − and δ ( π ( t )) = δ ( α n ) } . Proof.
We fix a natural number n . Choose an element t of A ( n − satisfying that δ ( π ( t )) = δ ( α n ).Let us see that if π ( t ) belongs to ˜ A , then the set A ( n ) contains π ( t ). Since α n < π ( t ) in S ( n − ,we have α n +1 < π ( π ( t )) in S ( n − and S ( n ) . By the construction of S ( n ) , we have π ( t ) < α n in S ( n ) . Hence the set A ( n ) contains π ( t ). Conversely, let t ′ be an element of A ( n ) . We denote by t the element of S ( n ) satisfying that π ( t ) = t ′ . We have to see that t belongs to A ( n − . In the sets S ( n ) and S ( n − , we have t < α n − . Moreover, since α n +1 < π ( π ( t )) holds in S ( n ) and S ( n − , weget α n < π ( t ) in S ( n − . Hence t belongs to A ( n − .Applying Proposition 3.11 repeatedly, we obtain the following: Corollary 3.12.
Let n be a natural number. Then A ( n ) is described as follows: A ( n ) = { π n ( t ) | t ∈ A (0) , δ ( π m ( t )) = δ ( α m ) and π m ( t ) ∈ ˜ A for all m with 0 ≤ m ≤ n } . Proposition 3.13.
Let n be a non-negative integer. Then the set A ( n ) does not contain an element t which is of the form t = π m (0 Ai ) with a non-negative integer m for m ≤ n . Proof.
For every non-negative integer n , the set A ( n ) consists of some elements of ˜ A . Hence sets A ( n ) do not contain the element π − (0 Ai ) which is an element of ˜ B . We show the assertion by inductionon n . The case n = 0 is obvious. For a natural number n , assume that the set A ( n − contain noelement π m (0 Ai ) with m ≤ n −
1. We consider the set A ( n ) . Suppose that A ( n ) contains an element π m (0 Ai ) for m ≤ n . Then by Proposition 3.11, π m − (0 Ai ) belongs to A ( n − . This contradicts withthe hypothesis of induction.N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 15 Proposition 3.14.
For every small modification by 0 Ai and 1 Bj with m < i ≤ n , there exists anon-negative integer a = a ( i, j ) such that A ( a ) = ∅ . Proof.
In the case of i = n and 1 ≤ j ≤ n , we immediately have A (0) = ∅ . Let us see the othercases. By the hypothesis m < i ≤ n , the map π sends 0 Ai + m to 1 Bj in S ( n ) for all non-negativeintegers n . Let m be the minimum natural number satisfying π m (0 Ai ) = 1 Am . If there exists anatural number n such that A ( n − = { Ai + m } for n < m , we have then A ( n ) = ∅ by definition of A ( n ) . Assume that sets A (0) , . . . , A ( m − are not empty. Let us consider the set A ( m ) . Note that1 Am is the maximum element of the set { t ∈ A | δ ( t ) = 1 } . Proposition 3.13 induces that everyelement t of A ( m − satisfies that δ ( π ( t )) = 0 or π ( t ) = 1 Bj , whence we see A ( m ) = ∅ .Thus by a small modification π by 0 Ai and 1 Bj , we obtain the non-negative integer a = a ( i, j )satisfying that A ( a ) = ∅ . By the ABS ( S ( a ) , δ, π ), we obtain the set B (0) = { B , . . . , Bj − } ∪ I ,where I is the subset of S ( a ) consisting of elements α n = π n (0 Ai ) satisfying that 1 Bj < α n and α n +1 < π (1 Bj ). Here, we show some properties of sets B (0) , B (1) , . . . in Lemma 3.15, Proposition 3.16,Proposition 3.17, Lemma 3.18 and Lemma 3.19 to see that there exists a non-negative integer n such that B ( n ) = ∅ in Proposition 3.20. We suppose that if n = m + 1 and m = n + 1, then( i, j ) = ( n , m ). Lemma 3.15.
Let n ′ be a non-negative integer with n ′ ≥ a . Set n = n ′ − a . Let Σ ′ ( n ) be the setconsisting of elements t of ˜ S satisfying that t < t ′ and π ( t ′ ) < π ( t ) in S ( n ′ ) for some elements t ′ of S ( n ′ ) with δ ( t ) = δ ( t ′ ). If B ( n ) is not empty, we have then Σ ′ ( n ) = { β n } , where β n = π n (1 Bj ).Moreover, if B ( n ) = ∅ , then Σ ′ ( n ) = ∅ . Proof.
A proof is given by the same way as Lemma 3.10.
Proposition 3.16.
Set β m = π m (1 Bj ) for all non-negative integers m . Let n be a natural number.The set B ( n ) is obtained by B ( n ) = { π ( t ) ∈ S ( a + n ) | t ∈ B ( n − and δ ( π ( t )) = δ ( β n ) } . Moreover, this set is described as B ( n ) = { π n ( t ) | t ∈ B (0) and δ ( π m ( t )) = δ ( β m ) for all m with 0 ≤ m ≤ n } . Proof.
A proof is given by the same way as Proposition 3.11.
Proposition 3.17.
Let n be a non-negative integer. If the set B ( n ) is a subset of B , then B ( n ) doesnot contain an element t which is of the form t = π m (1 Bj ) with a non-negative integer m for m ≤ n . Proof.
A proof is given by the same way as Proposition 3.13.
Lemma 3.18.
If 1 ≤ j ≤ n , then all elements t of I satisfy that δ ( π ( t )) = 0. Proof. If m and n satisfy n − m > m , then δ ( π ( t )) = 0 holds for all elements t of A with δ ( t ) = 1. It suffices to see the case of n − m < m . Assume that there exists an element t of I satisfying δ ( π ( t )) = 1. We recall that there exists a natural number n such that t = π n (0 Ai ). On theother hand, we can denote by t = 1 Ar this element t with a natural number r . Put r ′ = r + n − m .We have then π (1 Ar ) = 1 Ar ′ . By the definition of I , the set A ( n − contains the inverse image of 1 Bj .By construction, we have |A ( m ) | < n − m for all m . In S ( n − , the number of elements between 1 Ar and 1 Bj is greater than |A ( n − | , whence we have 1 Ar < Bj in S ( n ) . This contradict with 1 Ar ∈ I .6 NOBUHIRO HIGUCHI Lemma 3.19.
Let n be a natural number with n >
1. Suppose 1 ≤ j ≤ n . Then B ( n ) is a subsetof B . Proof.
First, let us see that sets B ( n ) do not contain 0 Ai for all n . Assume that a set B ( n − containsthe inverse image of 0 Ai . Note that the inverse image of 0 Ai is obtained by 0 Bj + m in this hypothesis.We have then δ ( π n − (1 Bj )) = 0 by the definition of B ( n − . The condition λ < / δ ( π n (1 Bj )) = 1. It implies that the element 0 Ai does not belong to B ( n ) .To show the statement of this lemma, let us consider two cases depending on the value of δ ( π (1 Bj )). On the one hand, suppose that δ ( π (1 Bj )) = 1, and let us see that B ( n ) consists of someelements of B for n ≥
1. We have B (1) = { π ( t ) | t ∈ B (0) with δ ( π ( t )) = 1 } in this assumption. Forelements t of I , all elements π ( t ) do not belong to B (1) since δ ( π ( t )) = 0 holds for all elements t of˜ A with δ ( t ) = 1. It induces that B (1) is a subset of B , whence we see that B ( n ) ⊂ B for all naturalnumbers in this case.On the other hand, suppose δ ( π (1 Bj )) = 0. Let Λ be the subset of B (1) consisting of elements t which belong to ˜ B . Then B (1) is obtained by the union of Λ and π ( I ). By the condition of m and n , for all elements t of Λ, we have δ ( π ( t )) = 1. Moreover, by Lemma 3.18, we have δ ( π ( t )) = 0 forall elements t of π ( I ). Thus we see that B (2) is a subset of B since δ ( π (1 Bj )) = 1 in this hypothesis.Hence B ( n ) is a subset of ˜ B for every natural number n with n > S ′ of S are obtained by the combinatorial method. Proposition 3.20.
For the small modification by 0 Ai and 1 Bj with m < i ≤ n , there exists anon-negative integer b = b ( i, j ) such that B ( b ) = ∅ . Proof.
First, let us consider the case 1 ≤ j ≤ n . If j = 1, then we have B (0) = I and B (1) = ∅ .For j >
1, we have B (0) = { B , . . . , Bj − } ∪ I . Let m be the minimum natural number satisfying π m (1 Bj ) = 0 Bm +1 . If m = 1, then B (1) = π ( I ). Since δ ( π (1 Bj )) = 1 in this case and Lemma 3.18,we have B (2) = ∅ . Let us see the case m >
1. Suppose that there exists a natural number n ,with n < m , such that B ( n ) = { Bj + m } , i.e., the set B ( n ) consists of an element the inverse imageof 0 Ai . By the proof of Lemma 3.19, the element 0 Ai does not belong to B ( n +1) , whence we have B ( n +1) = ∅ . Assume that B (0) , . . . , B ( m − are non-empty sets. Note that 0 Bm +1 is the minimumelement of all elements t of ˜ B satisfying δ ( t ) = 0. For an element t of B ( m − , if π ( t ) belongs to˜ B , then δ ( π ( t )) = 1 holds. Moreover, it follows from Lemma 3.19 that B ( m ) does not contain anelement of ˜ A . Therefore we have B ( m ) = ∅ .Next, let us see the case n < j ≤ m . We divide the proof into two cases depending on whether I is empty. First, suppose I = ∅ . For the minimum non-negative integer n satisfying that π n (1 Bj ) =1 B , we have B ( n ) = ∅ . In fact, if the set B ( n − is not empty, then B ( n − = { Ai } holds since π n − (1 Bj ) = 0 Bm +1 . If δ ( π (0 Ai )) = 1, then I contains π (0 Ai ) since A (0) = { Ai +1 , . . . , Am + n } containsthe inverse image of 1 Bj . This contradicts with the assumption. Hence we have δ ( π (0 Ai )) = 0, andthen B ( n ) = ∅ holds.Next, suppose I = ∅ . If j > n + 1, then the non-negative integer b is obtained by the number n satisfying that π n (1 Bj ) = 1 Bn +1 . In fact, B ( n − consists of some elements t of ˜ A since π n − (1 Bj ) = 1 B is the minimum element of ˜ B . It is clear that δ ( π ( t )) = 0 for all elements t of B ( n − , and we havethen B ( n ) = ∅ . If j = n + 1, then the non-negative integer b is obtained by the number n satisfying π n (1 Bj ) = 1 Bn +2 .N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 17For the Newton polygon ξ , we now suppose n = m + 1 and m = n + 1 with m > n >
0. If m < i ≤ n and n < j ≤ m , then i and j must be i = m + 1 and j = m . Weconstruct the specialization by a well-known method using binary expansions for such ABS’s. Forthe ABS S = ( ˜ S, δ, π ) corresponding to N ξ , we have then the following diagram of S :1 A · · · A Am +1 { { B · · · B A · · · A v v Bm > > B · · · B . Note that the diagram satisfies that δ ( t ) = δ ( π − ( t )) for all elements t except for these two elements.Let π be the small modification by 0 Am +1 and 1 Bm , and we obtain the admissible ABS S ′ = ( ˜ S ′ , δ, π ).By the construction of S ′ , we have δ ( t ) = δ ( π ( t )) for all elements of ˜ S ′ . Hence we obtain binaryexpansions of elements by b ( t ) = 0 . · · · if δ ( t ) = 0 and b ( t ) = 0 . · · · otherwise. Therefore,the ABS S ′ is associated to a DM mN , with m = m + m . Note that this ABS satisfies that ℓ ( S ′ ) < ℓ ( S ) −
1. In this case, by the small modification, we have B (0) = { A , . . . , Am , B , . . . , Bm } .For two elements t and t ′ of B (0) and for every non-negative integer n , we have δ ( π n ( t )) = δ ( π n ( t ′ )).Hence there exists no non-negative integer n satisfying B ( n ) = ∅ .Let S be the ABS associated to N ξ with a Newton polygon ξ = ( m , n )+( m , n ) satisfying that λ < / < λ . For the case ξ = ( m , m + 1) + ( m , m − S obtained by exchanging 0 Am +1 and 1 Bm is associated to the DM ( m + m ) N , . Moreover, for theother cases, we obtain specializations by full modifications. From now on, for a small modification π by 0 Ai and 1 Bj , we denote by a and b the minimum non-negative integers satisfying that A ( a ) = ∅ and B ( b ) = ∅ . The specialization S ′ obtained by exchanging 0 Ai and 1 Bj is equal to the ABS ( S ( a + b ) , δ, π ). In Example 3.9, we introduced an example of full modifications by 0 Ai and 1 Bj with m < i ≤ n and 1 ≤ j ≤ n . Here, let us see examples of the case n < j ≤ m , which are examples of thelatter part of the proof of Proposition 3.20. Example 3.21.
Let N = N , ⊕ N , . For the ABS S = ( ˜ S, δ, π ) associated to N , we constructthe small modification π by 0 A and 1 B . Then we get A (0) = { A , A , A } and a = 2. The ABS( S (2) , δ, π ) is described as( S (2) , δ, π ) : 1 A A (cid:10) (cid:10) A (cid:10) (cid:10) A B ◦ A (cid:127) (cid:127) A (cid:127) (cid:127) B × : : B × : : B × : : A x x A x x B × ? ? A w w B w w B w w B w w . Then we have I = ∅ , and sets B ( n ) are given by B (0) = { B , B , B , B } , B (1) = { A , B , B } , B (2) = { B , B } , B (3) = { A } , B (4) = ∅ . The ABS S ′ is obtained by the following: S ′ = 1 A A A A B A B B A B A A B B B A B . Example 3.22.
Let N = N , ⊕ N , . For the ABS S = ( ˜ S, δ, π ) associated to N , we constructthe small modification π by 0 A and 1 B . Then we get A (0) = { A , A , A , A , A } and a = 1. TheABS ( S (1) , δ, π ) is described as follows:( S (1) , δ, π ) : 1 A A (cid:10) (cid:10) B ◦ A (cid:4) (cid:4) A (cid:4) (cid:4) A (cid:4) (cid:4) A × : : B × : : B × : : B × : : A x x A x x A x x B ? ? B w w B w w B w w . Then I = { A } , and we have sets B (0) = { A , B , B , B } , B (1) = { A , A , B } , B (2) = { A , B } , B (3) = ∅ . The ABS S ′ is obtained by the following: S ′ = 1 A A B A A B A A B B A B B A A B B . Let ξ be a Newton polygon consisting of two segments satisfying λ < / < λ . In this section, forthe arrowed binary sequence S associated to the minimal DM N ξ , we characterize specializations S ′ of S satisfying ℓ ( S ′ ) = ℓ ( S ) −
1, i.e., we classify boundary components of central streams.Moreover, in Section 4.2, we show some properties of generic specializations. We will give a proofof Theorem 1.3 in Section 5 using these properties.
Now, we state the first result (Theorem 4.1). Let ξ = ( m , n ) + ( m , n ) be a Newton polygonwith λ < / < λ , and let N ξ be the minimal DM associated to ξ . Let S = A ⊕ B be the ABScorresponding to N ξ , where A (resp. B ) is the ABS corresponding to N m ,n (resp. N m ,n ). Put( ˜ S, δ, π ) = S . Set h = m + n and h = m + n . We construct the small modification π by 0 Ai with m < i ≤ n and 1 Bj with 1 ≤ j ≤ m . Then for the notation of Section 3 we obtain sets A ( n ) for 0 ≤ n ≤ a , sets B ( n ) for 0 ≤ n ≤ b and ABS’s ( S ( n ) , δ, π ) for 0 ≤ n ≤ a + b , where a and b arethe smallest non-negative integers satisfying that A ( a ) = ∅ and B ( b ) = ∅ . The main result of thissection is Theorem 4.1.
Let ξ = ( m , n ) + ( m , n ) be a Newton polygon with λ < / < λ . Let S = ( ˜ S, δ, π ) be the ABS associated to the DM N ξ . Let S ′ = ( ˜ S ′ , δ, π ) be the ABS obtained byexchanging 0 Ai and 1 Bj . Then ℓ ( S ′ ) = ℓ ( S ) − n such that A ( n ) contains the inverse image of 1 Bj or B ( n ) contains the inverse image of 0 Ai for the small modification π .This theorem gives a classification of generic specializations of H ( ξ ). To show the above, wedivide the problem into three cases depending on conditions of i and j as follows. Definition 4.2.
For the ABS S = A ⊕ B associated to the minimal DM N ξ with the Newtonpolygon ξ = ( m , n )+( m , n ), we denote by S ′ = S ′ ( i, j ) the specialization obtained by exchangingN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 190 Ai and 1 Bj . We define sets H ( S ) = { S ′ ( i, j ) | m < i ≤ n and 1 ≤ j ≤ n } , H ( S ) = { S ′ ( i, j ) | m < i ≤ n and n < j ≤ m } , H ( S ) = { S ′ ( i, j ) | n < i ≤ h and n < j ≤ m } . Let T be the ABS associated to a DM N , and let T D be the ABS associated to the dual N D of N . We call this ABS dual ABS of T . For the above notation, we will give a concrete conditionof i and j satisfying that the exchange of 0 Ai and 1 Bj is a good. Theorem 4.1 follows from thisproposition: Proposition 4.3.
The following holds:(i) For S ′ ∈ H ( S ), the formula ℓ ( S ′ ) = ℓ ( S ) − n satisfying that A ( n ) contains 0 Ai + m or B ( n ) contains 0 Bj + m .(ii) For S ′ ∈ H ( S ), we have ℓ ( S ′ ) < ℓ ( S ) − S ′ ∈ H ( S ), the formula ℓ ( S ′ ) = ℓ ( S ) − S D and S ′ D satisfy ℓ ( S ′ D ) = ℓ ( S D ) − S ′ of H ( S ), we immediately see that 0 Ai + m and 0 Bj + m are inverseimages of 1 Bj and 0 Ai respectively.If the Newton polygon ξ = ( m , n ) + ( m , n ) satisfies that n = m + 1 and m = n + 1 with m > n >
0, then the specialization S ′ obtained by exchanging 0 An and 1 Bm corresponds tothe DM mN , with m = m + m . Clearly this S ′ satisfies ℓ ( S ′ ) < ℓ ( S ) −
1. In this case, theset A (0) contains the inverse image of 1 Bm for the small modification π by 0 An and 1 Bm . We mayassume that every specialization S ′ is obtained by the full modification ( S ( a + b ) , δ, π ).First, we will show (i) and (ii) of Proposition 4.3. For the ABS ( ˜ S, δ, π ) associated to N ξ , fixelements 0 Ai and 1 Bj with m < i ≤ n . Let π be the small modification by 0 Ai and 1 Bj . Let a (resp. b ) denote the smallest non-negative integer such that A ( a ) = ∅ (resp. B ( b ) = ∅ ). We introduce somedefinitions to calculate the length of the specialization S ′ = ( S ( a + b ) , δ, π ). For simplicity, we oftenwrite ℓ ( S ( n ) ) for the length of the ABS ( S ( n ) , δ, π ). Notation 4.4.
For non-negative integers n with n < a , we define d A ( n ) by d A ( n ) = |A ( n ) | − |A ( n +1) | . Moreover, we define ∆ ℓ A ( n ) by ∆ ℓ A ( n ) = ℓ ( S ( n +1) ) − ℓ ( S ( n ) ) . Put ∆ ℓ A = P n ∆ ℓ A ( n ). Notation 4.5.
Let n ′ be a non-negative integer with n ′ ≥ a . Put n = n ′ − a . We define d B ( n ) by d B ( n ) = |B ( n ) | − |B ( n +1) | . Moreover, we define ∆ ℓ B ( n ) by ∆ ℓ B ( n ) = ℓ ( S ( n ′ +1) ) − ℓ ( S ( n ′ ) ) . Put ∆ ℓ B = P n ∆ ℓ B ( n ).0 NOBUHIRO HIGUCHI Example 4.6.
Let N = N , ⊕ N , . In Example 3.9, we constructed the small modification byelements 0 A and 1 B . By this small modification, the length of the ABS’s decrease by four from S to S (0) . We have ∆ ℓ A = 1 and ∆ ℓ B = 2. Thus the length is increased by three from S (0) to S ′ , andeventually we see ℓ ( S ′ ) = ℓ ( S ) − d A ( n ) and d B ( n ), we obtain Lemma 4.7 and Lemma 4.8 which areused for evaluating values ∆ ℓ A and ∆ ℓ B . Recall that I is the subset of B (0) consisting of elements t of ˜ A . Lemma 4.7.
Let S ′ ∈ H ( S ). The following are true:(1) P n d A ( n ) = n − i ,(2) P n d B ( n ) = | I | + j − Proof.
Clearly P n d A ( n ) is given by |A (0) | − |A ( a ) | . As |A (0) | = n − i and |A ( a ) | = 0, we obtainthe desired value. Similarly, we obtain (2) since |B (0) | = | I | + j − |B ( b ) | = 0. Lemma 4.8.
Let S ′ ∈ H ( S ). The following are true:(1) P n d A ( n ) = h − i ,(2) P n d B ( n ) = | I | + j − Proof.
Note that we have A (0) = { Ai +1 , . . . , Ah } and B (0) = I ∪ { B , . . . , Bj − } in this case. A proofis given by the same way as Lemma 4.7. Notation 4.9.
For an element t of S ( n ) with δ ( t ) = 1, we define ℓ ( t, n ) by the number of elements t ′ satisfying that t ′ < t and δ ( t ′ ) = 0 in S ( n ) . For instance, the sum P t ℓ ( t, n ) is equal to the lengthof the ABS ( S ( n ) , δ, π ).We will give a criterion of generic specializations in Proposition 4.3 and Theorem 4.1 by compar-ing values of d A ( n ) and ∆ ℓ A ( n ), or d B ( n ) and ∆ ℓ B ( n ) using Proposition 4.11 and Proposition 4.12below. Lemma 4.10.
Let n be a non-negative integer. Put α = π n +1 (0 Ai ) and β = π n +1 (1 Bj ). Thefollowing holds: | ∆ ℓ A ( n ) | = { t ∈ A ( n ) | δ ( π ( t )) = δ ( α ) } , (8) | ∆ ℓ B ( n ) | = { t ∈ B ( n ) | δ ( π ( t )) = δ ( β ) } . (9)Concretely, if δ ( α ) = 0 (resp. δ ( α ) = 1), we have then ∆ ℓ A ( n ) ≤ ℓ A ( n ) ≥ δ ( β ) = 0 (resp. δ ( β ) = 1), we have then ∆ ℓ B ( n ) ≥ ℓ B ( n ) ≤ Proof.
We fix a non-negative integer n , and let us show that the equation (8) holds. In the sameway, we can obtain the equation (9). We divide the proof into two cases depending on the value of δ ( α ). First, suppose δ ( α ) = 0. If π ( A ( n ) ) does not contain 1 Bj , then it follows from Proposition 3.13that all elements t of π ( A ( n ) ) satisfy δ ( t ) = 0, and hence we have d A ( n ) = 0. Moreover, in thiscase ∆ ℓ A ( n ) = 0 holds. If π ( A ( n ) ) contains 1 Bj , then all elements t of π ( A ( n ) ) satisfy δ ( t ) = 0except for 1 Bj , whence we have d A ( n ) = 1. Then ℓ (1 Bj , n + 1) = ℓ (1 Bj , n ) − ℓ ( t, n + 1) = ℓ ( t, n ) holds for the other elements, whence we have ∆ ℓ A ( n ) = −
1. Next, let us seeN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 21the case of δ ( α ) = 1. By the construction of S ( n +1) , it is clear that ℓ ( α, n + 1) = ℓ ( α, n ) + r , where r = { t ∈ A ( n ) | δ ( π ( t )) = δ ( α ) } . Moreover, ℓ ( t, n + 1) = ℓ ( t, n ) holds for the other elements t .Clearly we have d B ( n ) = r . Hence we get desired equality for the case of δ ( α ) = 1. This completesthe proof. Proposition 4.11.
For all non-negative integers n , an inequality ∆ ℓ A ( n ) ≤ d A ( n ) holds. Theequality holds for all n if and only if A ( n ) do not contain the inverse image of 1 Bj for all n . Proof.
By the condition m < i ≤ n , in ABS’s ( S ( n ) , δ, π ), the inverse image of 1 Bj is obtained by0 Ai + m . We fix a non-negative integer n with n < a . Put α = π n +1 (0 Ai ). In the ordered set S ( n +1) ,the element α is located in the right of elements of π ( A ( n ) ). Let us suppose that A ( n ) does notcontain 0 Ai + m . This assumption implies that π ( A ( n ) ) contains no element of B . If δ ( α ) = 0, thenall elements t of A ( n ) satisfy δ ( π ( t )) = 0. Hence we have ∆ ℓ A ( n ) = 0, and then d A ( n ) = 0 holds. If δ ( α ) = 1, by Proposition 3.11, the set A ( n +1) is obtained by π ( A ( n ) ) \ Ξ, whereΞ = { π ( t ) | t ∈ A ( n ) and δ ( π ( t )) = δ ( α ) } . We immediately obtain | Ξ | = d A ( n ). Lemma 4.10 concludes that ∆ ℓ A ( n ) = d A ( n ).Assume that there exists a non-negative integer n such that A ( n ) contains 0 Ai + m . We divide theproof into two cases depending on the value of δ ( α ). First, if δ ( α ) = 0, it follows from ℓ (1 Bj , n + 1) = ℓ (1 Bj , n ) − ℓ A ( n ) = −
1. As A ( n +1) is obtained by A ( n +1) = π ( A ( n ) ) \ { Bj } , we have d A ( n ) = 1. Hence we get ∆ ℓ A ( n ) < d A ( n ). Next, if δ ( α ) = 1, we obtain the set A ( n +1) by π ( A ( n ) ) \ Ξ ′ , where Ξ ′ = { π ( t ) | t ∈ A ( n ) , with δ ( π ( t )) = 0 or π ( t ) ∈ ˜ B } . It is clear that Ξ ′ contains 1 Bj in this hypothesis. We have d A ( n ) = | Ξ ′ | . On the other hand, since δ (1 Bj ) = δ ( α ), we have ℓ ( α, n + 1) = ℓ ( α, n ) + ( | Ξ ′ | − ℓ A ( n ) = | Ξ ′ | −
1. Hencewe get ∆ ℓ A ( n ) < d A ( n ). Proposition 4.12.
For all non-negative integers n , an inequality ∆ ℓ B ( n ) ≤ d B ( n ) holds. Moreover,for 1 ≤ j ≤ n , the equality holds for all n if and only if(1) B ( n ) do not contain the inverse image of 0 Ai for all n , and(2) I = ∅ . Proof.
For all j , the inequality follows from Lemma 4.10. To see the latter part, we treat the caseof 1 ≤ j ≤ n . In this hypothesis, in ABS’s ( S ( n ) , δ, π ), the inverse image of 0 Ai is obtained by0 Bj + m . If sets B ( n ) do not contain 0 Bj + m for all n and I = ∅ , then we can show that the equality∆ ℓ B ( n ) = d B ( n ) holds in the same way as Proposition 4.11.Let us see the converse. Put β n = π n (1 Bj ) for non-negative integers n . We assume that B ( n ) contains 0 Bj + m for a non-negative integer n . By the condition n /h < / h = m + n ,we have δ ( t ) = 0 and δ ( π ( t )) = 1 for all elements t of B ( n ) except for 0 Bj + m . Moreover, we have δ ( β n ) = 0 and δ ( β n +1 ) = 1. In the ABS ( S ( a + n +1) , δ, π ) we have β n +1 < π ( t min ), where t min is theminimum element of B ( n ) . We have then ∆ ℓ B ( n ) = − ℓ ( β n +1 , n + 1) = ℓ ( β n +1 , n ) − ℓ ( t, n + 1) = ℓ ( t, n ) for the other elements t of S ( a + n +1) . On the other hand, we have d B ( n ) = 1. Infact, B ( n +1) is given by B ( n +1) = π ( B ( n ) ) \ { Ai } .2 NOBUHIRO HIGUCHINext, assume I = ∅ . We divide the proof into two cases depending on values of δ ( β ). If δ ( β ) = 0, then the set B (1) is the union of Λ and π ( I ), where Λ is the subset of B (1) consisting ofelements t of B (1) satisfying t ∈ B . Note that δ ( β ) = 1 in this hypothesis. We have δ ( π ( t )) = 1 forevery element t of Λ. Moreover, Lemma 3.18 implies that δ ( π ( t )) = 0 for every element t of π ( I ).Hence we have ℓ ( β ,
2) = ℓ ( β , − | I | , and it implies that ∆ ℓ B (1) = −| I | . On the other hand, wehave d B (1) = | I | , whence ∆ ℓ B (1) < d B (1) holds. Let us suppose δ ( β ) = 1. In this case, we obtain∆ ℓ B (0) = −| I | . Since d B (0) = | I | , we have ∆ ℓ B (0) < d B (0).Thanks to the above propositions, we can prove Proposition 4.3. Proof of Proposition 4.3 (i).
We have ℓ (( S (0) , δ, π )) − ℓ ( S ) = − ( n − i + j ). Note that if thereexists no non-negative integer n such that A ( n ) contains the inverse image of 1 Bj , then I = ∅ holds. Furthermore, if there exists no non-negative integer n such that B ( n ) contains the inverseimage of 0 Ai , then Lemma 4.7, Proposition 4.11 and Proposition 4.12 imply that ∆ ℓ A = n − i and∆ ℓ B = j −
1. Hence we have ℓ ( S ′ ) − ℓ (( S (0) , δ, π )) = n − i + j −
1, and ℓ ( S ′ ) = ℓ ( S ) − n such that A ( n ) contains the inverse image of1 Bj or B ( n ) contains the inverse image of 0 Ai . If I = ∅ , then we have ∆ ℓ A < n − i or ∆ ℓ B < j − ℓ ( S ′ ) < ℓ ( S ) −
1. On the other hand, if I = ∅ , then we have ∆ ℓ A ≤ n − i − | I | .Moreover, by the proof of Proposition 4.12, as there exists a non-negative integer m such that∆ ℓ B ( m ) = −| I | , we have ∆ ℓ B < j −
1. Hence we have ℓ ( S ′ ) < ℓ ( S ) − Proof of Proposition 4.3 (ii).
In this case, the set A (0) is given by A (0) = { Ai +1 , . . . , Ah } . We have ℓ (( S (0) , δ, π )) − ℓ ( S ) = − ( h − i + j ). By the condition of i , the set A (0) contains 0 Ai + m which is theinverse image of 1 Bj . Hence we have ∆ ℓ A < h − i by Lemma 4.8 and Proposition 4.11. Moreover,we have ∆ ℓ B ≤ j − B ( n ) contains the element π n ( t ) for t ∈ I , then ∆ ℓ B ( n ) <
0. and hence ℓ ( S ′ ) < ℓ ( S ) − n = m + 1 and m = n + 1, for the small modification π by0 An and 1 Bm , the set A (0) contains the inverse image of 1 Bm .Let us classify specializations satisfying ℓ ( S ′ ) = ℓ ( S ) − S ′ ∈ H ( S ). We use the dualityto consider this case. Let N = N ξ be the DM with a Newton polygon ξ = ( m , n ) + ( m , n )satisfying λ < / < λ . Let N D be the dual of N . Then we have N D = N n ,m ⊕ N n ,m . Let S D be the ABS corresponding to N D . Note that ℓ ( S ) = ℓ ( S D ) follows from the definition of thelengths of ABS’s. Proof of Proposition 4.3 (iii).
Fix an ABS S ′ ∈ H ( S ). For the same notation as above, S D = B D ⊕ A D , where B D and A D correspond to N n ,m and N n ,m respectively. The elements 0 Ai and1 Bj correspond to 1 j ′ = 1 A D j ′ and 0 i ′ = 0 B D i ′ by the duality, with i ′ = h − j + 1 and j ′ = h − i + 1.Then S ′ D = ( S D ) ′ is the ABS obtained by the small modification by 0 i ′ and 1 j ′ in S D . Since S ′ D belongs to H ( S D ), the exchange of 0 i ′ and 1 j ′ in S D is good if and only if the small modificationby 0 i ′ and 1 j ′ satisfies the necessary and sufficient condition of Proposition 4.3 (i). As the smallmodification by 0 Ai and 1 Bj in S corresponds to the small modification by 0 i ′ and 1 j ′ in S D , theequality ℓ ( S ′ ) = ℓ ( S ) − ℓ ( S ′ D ) = ℓ ( S D ) − Proof of Theorem 4.1.
Proposition 4.3 (i) and (ii) imply that the statement of Theorem 4.1 holdsfor ABS’s S ′ of H ( S ) and H ( S ). Moreover, by the duality, Proposition 4.3 (iii) concludes thatN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 23 ℓ ( S ′ ) = ℓ ( S ) − n such that A ( n ) (resp. B ( n ) ) does not contain the inverse image of 1 Bj (resp. 0 Ai ) for S ′ ∈ H ( S ).We want to determine boundary components of H ( ξ ) for arbitrary Newton polygons ξ . If Con-jecture 4.13 is true, then the two segments case is essential for classification of boundary componentsof all central streams. Conjecture 4.13.
Let ξ be a Newton polygon of z segments. Let N ξ be the minimal DM of ξ .A specialization of N ξ is generic if and only if it is the direct sum of a generic specialization of N r ⊕ N r +1 and the minimal p -divisible group N ⊕ · · · ⊕ N r − ⊕ N r +2 ⊕ · · · ⊕ N z for a natural number r . Here, N i is the simple DM associated with the i -th segment of ξ . Here, we introduce some notation and properties of generic specializations, which are useful forshowing Theorem 1.3.Theorem 4.1 and Proposition 4.3 imply that it suffices to deal with ABS’s of H ( S ) to studyboundary components of central streams. From now on, in this paper, for elements 0 Ai and 1 Bj used in the construction of the small modification π , we make the assumption m < i ≤ n and1 ≤ j ≤ n . Note that for the good exchange of 0 Ai and 1 Bj , if a and b are positive, then non-negativeintegers a and b satisfy that π a (0 Ai ) = 1 Am and π b (1 Bj ) = 0 Bm +1 .By Theorem 4.3, if the exchange of 0 Ai and 1 Bj is good, then sets A ( n ) (resp. B ( n ) ) are subsetsof ˜ A (resp. ˜ B ) for all non-negative integers n , whence for the small modification by 0 Ai and 1 Bj , sets A ( n ) (resp. B ( n ) ) and the value ∆ ℓ A (resp. ∆ ℓ B ) do not depend on j (resp. i ). Thus we define thefollowing sets. Definition 4.14.
Let A = ( ˜ A, δ A , π A ) and B = ( ˜ B, δ B , π B ) be the ABS of N m ,n and N m ,n respectively. For the ABS S = A ⊕ B , which is associated to N ξ , let G be the subset of ˜ A × ˜ B consisting of pairs (0 Ai , Bj ) such that exchanges of 0 Ai and 1 Bj are good. By the above, this set isdescribed as G = C ′ × D ′ with C ′ ⊂ ˜ A and D ′ ⊂ ˜ B . Equivalently, we obtain a generic specializationof S if and only if we construct a small modification by an element of C ′ and an element of D ′ .Moreover, let C (resp. D ) be the subset of C ′ (resp. D ′ ) consisting of elements satisfying that A (0) (resp. B (0) ) is not empty.From now on, we fix the notation of C ′ , C, D ′ and D . We have C ′ \ C = { An } and D ′ \ D = { B } .We call the construction of a full modification using an element of C ′ and an element of D ′ goodexchange. Example 4.15.
Let N ξ = N , ⊕ N , . Then the set G = C ′ × D ′ is given by C ′ × D ′ = { (0 A , B ) , (0 A , B ) , (0 A , B ) , (0 A , B ) } . We have C ′ = { A , A } and D ′ = { B , B } .In Lemma 4.16 and Proposition 4.17, we give some properties of {A ( n ) } n =0 ,...,a and {B ( n ) } n =0 ,...,b for generic specializations. For the ABS S associated to N ξ , let ( S ( n ) , δ, π ) be ABS’s obtained byconstructing the full modification by 0 Ai ∈ C ′ and 1 Bj ∈ D ′ for n = 0 , . . . , a + b .4 NOBUHIRO HIGUCHI Lemma 4.16.
Let n be a non-negative integer. For a good exchange, if d A ( n ) > d B ( n ) > A ( n +1) (resp. B ( n +1) ) is not empty, then A ( n +1) (resp. B ( n +1) ) has the maximum element 1 Am (resp. the minimum element 0 Bm +1 ). Proof.
Let us see the case d A ( n ) >
0. We can show the case d B ( n ) > n . Put α = π n (0 Ai ). Clearly, if δ ( α ) = δ ( π ( α )) = 0, then δ ( π ( t )) = δ ( π ( α ))holds for all elements t of A ( n ) . Moreover, if δ ( α ) = 1, we have then δ ( π ( α )) = δ ( π ( t )) = 0 for allelements t of A ( n ) . It implies that d A ( n ) = 0 holds in these cases. Hence it suffices to see the casethat δ ( α ) = 0 and δ ( π ( α )) = 1. Then we have π ( A ( n ) ) = { Ax , . . . , Am , Am +1 , . . . , Ay } . It inducesthat the maximum element of A ( n +1) is 1 Am . Proposition 4.17.
For a good exchange, let n be a non-negative integer satisfying n < a , and let c be the natural number satisfying that π n (0 Ai ) = τ Ac with τ = 0 or 1. The set A ( n ) is obtained by A ( n ) = { τ Ac +1 , τ Ac +2 , . . . , τ Ac + u } , where u = |A ( n ) | . Similarly, for a non-negative integer n with n < b and the natural number c satisfying π n (1 Bj ) = τ Bc , the set B ( n ) is obtained by B ( n ) = { τ Bc − v , τ Bc − v +1 , . . . , τ Bc − } , where v = |B ( n ) | . Proof.
Let us see sets A ( n ) . Note that we have I = ∅ , and we can see the latter part in the sameway. We use induction on n . Clearly the statement holds for n = 0. For a natural number n ,suppose that A ( n − = { τ Ac +1 , . . . , τ Ac + u } with u = |A ( n − | . We have A ( n ) = π ( A ( n − ) \ T , with T = { π ( t ) | δ ( π ( t )) = δ ( π n (0 Ai )) for t ∈ A ( n − } . If T = ∅ , we clearly have the property for A ( n ) . Suppose that T = ∅ . By Lemma 4.16, we see A ( n ) = { Ac +1 , Ac +2 , . . . , Am − , Am } . This completes the proof. The main purpose of this section is to show Theorem 1.3. For a minimal DM N ξ and the ABS S associated to N ξ , we say a specialization N ′ of N ξ is generic if the specialization S ′ associated to N ′ satisfies ℓ ( S ′ ) = ℓ ( S ) −
1, and we denote by N − ξ this generic specialization of N ξ . We will showProposition 5.1 which is a key step of constructing specializations from a generic specialization of N ξ to a minimal DM N ζ with ζ ≺ ξ is saturated. The main result of this section is Proposition 5.1.
Let ξ = ( m , n ) + ( m , n ) be a Newton polygon satisfying λ < / < λ ,where λ = n / ( m + n ) and λ = n / ( m + n ). Assume that ξ is not (0 ,
1) + (1 , N ξ bethe DM of ξ . For every generic specialization N − ξ , there exist N −− ξ and N − ξ ′ satisfying(10) N −− ξ = N − ξ ′ ⊕ N ρ , where ρ = ( f, g ) is a Newton polygon and ρ is uniquely determined by ξ so that the area of theregion surrounded by ξ , ξ ′ and ρ is one. The diagram of these Newton polygons is described asN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 25either of the following:For the above statement, we will determine the diagram of Newton polygons in Proposition 5.2dividing the cases into the conditions of ξ . We fix notations as follows: Let ξ = ( m , n ) + ( m , n )be a Newton polygon satisfying that λ < / < λ . Set h = m + n and h = m + n . For theABS S = A ⊕ B associated to the DM N ξ = N m ,n ⊕ N m ,n , let S − denote the ABS obtained bya good exchange by 0 Ai and 1 Bj , and let N − ξ be the DM corresponding to S − . Recall that by thesmall modification by 0 Ai and 1 Bj , we obtain sets {A ( n ) } n =0 ,...,a and {B ( n ) } n =0 ,...,b , where a (resp. b )is the smallest non-negative integer satisfying that A ( a ) = ∅ (resp. B ( b ) = ∅ ). Note that if a and b are positive, then π a (0 Ai ) = 1 Am and π b (1 Bj ) = 0 Bm +1 . For the ABS S = ( ˜ S, δ, π ), the genericspecialization S − obtained by exchanging 0 Ai and 1 Bj is constructed by ABS’s { ( S ( n ) , δ, π ) } n =0 ,...,a + b .Since we can use the duality of DM ’s for S − ∈ H ( S ), it suffices to see the case that S − belongsto H ( S ). Hence we assume that m < i ≤ n and 1 ≤ j ≤ n .Theorem 1.3 is obtained by applying Proposition 5.1 inductively. In Proposition 5.2, we con-cretely give an operation to obtain N −− ξ satisfying the equality (10). Proposition 5.2.
Let ξ = ( m , n ) + ( m , n ) be a Newton polygon satisfying 0 < λ < / <λ <
1. Let N − ξ be any generic specialization of N ξ . Let S − be the ABS associated to N − ξ .(I) If n > m + 1, then we obtain the equality (10) for N −− ξ corresponding to S −− , where S −− is either of the following:(a) S −− is the specialization obtained by exchanging 0 Am +1 and 1 Am for S − , or(b) S −− is the specialization obtained by exchanging 0 Ai − and 1 Bj for S − .For these cases, the Newton polygon ξ ′ of (10) is of the form ξ ′ = ( m − f, n − g ) + ( m , n ).(II) If n = m + 1, then we obtain the equality (10) for N −− ξ corresponding to S −− , where S −− is any of the following:(c) S −− is the specialization obtained by exchanging 0 Bm +1 and 1 Bm for S − ,(d) S −− is the specialization obtained by exchanging 0 Ai and 1 Bj +1 for S − ,(e) S −− is the specialization obtained by exchanging 0 Ah and 1 Bn +1 for S − .For (c) and (d), the Newton polygon ξ ′ of (10) is of the form ξ ′ = ( m , n ) + ( m − f, n − g ),and for the case (e), we have ξ ′ = ( m − , n −
1) + ( m , n ).6 NOBUHIRO HIGUCHI Let us show Proposition 5.2 (I). Assume n > m + 1. By the construction of S − which is obtainedby a good exchange of 0 Ai ∈ C ′ and 1 Bj ∈ D ′ , we have 0 Ai − < Bj in this ABS; see Definition 4.14 forthe definition of sets C ′ and D ′ . In fact, it is clear that 0 Ai − < Bj in S (0) . Moreover, if 1 Bj < Ai − istrue in S − , then there exists a natural number n with n < a such that π n (0 Ai ) = 0 Ai − . Then A ( n − contains the inverse image of 1 Bj . This contradicts with ℓ ( S − ) = ℓ ( S ) −
1. Here, to treat the case(a) of Proposition 5.2, in Proposition 5.4, we will see that 0 Am +1 < Am holds in S − . The followingnotation is useful for showing some properties of ABS’s. Notation 5.3.
Let ( ˜
T , δ, π ) be an ABS. For an element t of ˜ T , we often express a subset { π m ( t ) | ≤ m ≤ n } of ˜ T as t → π ( t ) → · · · → π n ( t ) , and we call this diagram a path . Proposition 5.4.
We have 0 Am +1 < Am in the ABS S − obtained by a good exchange of 0 Ai ∈ C and 1 Bj ∈ D ′ . Moreover, there exists no non-negative integer n such that π n (0 Ai ) = 0 Am +1 with n ≤ a . Proof.
By the condition n − m >
1, we have h >
2. First, to show the former statement, let ussee that the non-negative integer a is not greater than h −
2. For a good exchange, the non-negativeinteger a satisfies that π a (0 Ai ) = 1 Am . We clearly have a < h . If a = h −
1, we have then i = h .This contradicts with the condition of the natural number i . Hence we have a ≤ h − A which is associated to the simple DM N m ,n , binary expansions of 1 Am and0 Am +1 are given by b (1 Am ) = 0 .b b · · · b h − · · · , (11) b (0 Am +1 ) = 0 .b b · · · b h − · · · . (12)For elements 0 Ai and 0 Ai +1 of S − , we have two paths as follows:0 Ai π −→ · · · π −→ Am , (13) 0 Ai +1 π −→ · · · π −→ Am +1 . (14)It is clear that 0 Ai +1 belongs to A (0) . Moreover, the above binary expansions and Corollary 3.12induce that π n (0 Ai +1 ) belongs to the set π ( A ( n − ) for every natural number n with n ≤ a . Inparticular, we apply this property for n = a , and we have 0 Am +1 ∈ π ( A ( a − ). By the constructionof S ( a ) , we obtain 0 Am +1 < Am in ( S ( a ) , δ, π ) and S − .Let us see the latter statement. Suppose that there exists a non-negative integer n with n ≤ a satisfying that π n (0 Ai ) = 0 Am +1 . If this hypothesis leads 1 Am < Am +1 , then we have a contradictionwith the former statement. Suppose that 0 Am +1 < Am in S − . Then the set π ( A ( a − ) contains theelement π n (0 Ai ). It implies that the set A ( a − contains the element π n − (0 Ai ). Proposition 3.13concludes that this is a contradiction, and hence 1 Am < Am +1 holds.In the same way, we obtain the following assertion which is used in the proof of Proposition 5.2(II).N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 27 Proposition 5.5.
For a good exchange of 0 Ai ∈ C ′ and 1 Bj ∈ D , we have 0 Bm +1 < Bm in the ABS S − . Moreover, there exists no non-negative integer n such that π n (1 Bj ) = 1 Bm with n ≤ b . Notation 5.6.
For the ABS A associated to N m ,n , we define sub-paths P and Q of A by thefollowing: P : 1 Am → Ah → · · · → A m +1 ,Q : 0 Am +1 → A → · · · → A m . Clearly A is a disjoint union of P and Q as sets.The above paths P and Q are useful. For instance, in the case (a) of Proposition 5.2, for theABS S ρ = ( ˜ S ρ , δ ρ , π ρ ) which is associated to N ρ of Proposition 5.1, the set ˜ S ρ consists of all elementsof P . Moreover, the path Q has the following property: Lemma 5.7.
The set C is contained in Q . Proof.
Suppose that there exists an element 0 Ai of C which belongs to P . Then there exists anatural number n such that π n (0 Ai ) = 0 Am +1 with n < a . This contradicts with the latter statementof Proposition 5.4.Here, let us consider the construction of the ABS obtained by (a) or (b) of Proposition 5.2. TheABS S − obtained by the small modification by 0 Ai and 1 Bj is described as the following diagram: • (cid:15) (cid:15) · · · o o Am o o A m o o · · · o o Ai o o Bj + m o o · · · o o • / / · · · / / A m +1 / / Am +1 / / · · · / / Ai + m / / Bj / / · · · elements of B O O First, let us consider the case (a). By constructing the small modification by 0 Am +1 and 1 Am ,images of 0 A m +1 and 0 A m are switched, and we obtain the ABS which consists of two componentsas follows: • (cid:15) (cid:15) · · · o o Am o o A m (cid:15) (cid:15) × o o · · · o o Ai o o Bj + m o o · · · o o • / / · · · / / A m +1 O O × / / Am +1 / / · · · / / Ai + m / / Bj / / · · · elements of B O O The former component consists of all elements of P . The DM corresponding to this componentis described as N ρ with a Newton polygon ρ = ( f, g ). Since this component coincides with thecomponent obtained from A by applying [1, Lemma 5.6] to the adjacent 1 Am Am +1 , we have f n − gm = 1. Next, let us see the case (b). By constructing the small modification by 0 Ai − and 1 Bj , images of 0 Ai − m and 0 Ai + m are switched, and we obtain the ABS which consists of twocomponents: • (cid:15) (cid:15) · · · o o Ai − o o Ai − m (cid:15) (cid:15) × o o · · · o o • o o • / / · · · / / Ai + m O O × / / Bj elements of B / / · · · / / Ai O O π n (0 Ai ) for all non-negative integers n with n ≤ a . Theformer component corresponds to the DM N ρ with a Newton polygon ρ = ( f, g ). Since thiscomponent coincides with the component obtained from A by applying [1, Lemma 5.6] to theadjacent 0 Ai − Ai , we have f n − gm = 1.Hence for the cases (a) and (b), the ABS S −− has two components, where one is associated tothe DM N ρ . We will show that for the other component, there exists a Newton polygon ξ ′ suchthat this component corresponds to the DM N − ξ ′ satisfying (10) of Proposition 5.2. Definition 5.8.
We define C (resp. C ) to be the subset of C ′ consisting of elements 0 Ai satisfyingthat for N − ξ obtained by a small modification by 0 Ai and 1 Bj ∈ D ′ , we have the equality (10) by thecase (a) (resp. (b)) of Proposition 5.2.We denote by S ′′ the ABS obtained by the small modification by 0 Am +1 and 1 Am or the smallmodification by 0 Ai − and 1 Bj for S − . By the above, we see that the ABS S ′′ consists of twocomponents, and a component of S ′′ is associated to N ρ . Let Ψ be the other component of S ′′ . Toshow C ∪ C = C ′ , we give a condition that an element t of C ′ belongs to C or C in Proposition 5.9.In the proof of this proposition, we give a method to determine the structure of the ABS Ψ. Proposition 5.9.
Let S − be the generic specialization obtained by a small modification by 0 Ai and1 Bj . If Ψ contains no element t satisfying that 0 Am +1 < t < Am (resp. 0 Ai − < t < Bj ) in S − , then0 Ai belongs to C (resp. C ). Proof.
Let us see the case that we construct S ′′ by the small modification by 0 Am +1 and 1 Am . Theother case ( S ′′ is constructed by the small modification by 0 Ai − and 1 Bj ) is shown by the sameway. For the ABS S − = ( ˜ S ′ , δ, π ), we construct the small modification π ′′ by 0 Am +1 and 1 Am ,and we obtain the admissible ABS S ′′ = ( ˜ S ′′ , δ, π ′′ ) consisting of two components. A componentcorresponds to N ρ with ρ = ( f, g ) satisfying f n − gm = 1. Let N be the DM associated to theother component Ψ. To see that there exists a Newton polygon ξ ′ such that N = N − ξ ′ , we considerthe small modification by 1 Bj and 0 Ai in S ′′ . Let χ be the small modification by 1 Bj and 0 Ai , and weobtain the map χ on ˜Ψ. For the ordered set { t < · · · < t h ′ } of Ψ, put Ψ (0) = { t ′ < ′ · · · < ′ t ′ h ′ } ,where if 0 Ai and 1 Bj are i ′ -th element and j ′ -th element of ˜Ψ respectively, then for s = ( i ′ , j ′ )transposition, we set t ′ z = t s ( z ) . We have the ABS (Ψ (0) , δ, χ ). Here, we define sets D (0) = { t ∈ Ψ (0) | Ai < t and χ ( t ) < χ (0 Ai ) in Ψ (0) with δ ( t ) = 0 } , E (0) = { t ∈ Ψ (0) | t < Bj and χ (1 Bj ) < χ ( t ) in Ψ (0) with δ ( t ) = 1 } , Put α n = χ n (0 Ai ) and β n = χ n (1 Bj ) for all non-negative integers n . For sets Ψ (0) , . . . , Ψ ( n − andsets D (0) , . . . , D ( n − with a natural number n , let Ψ ( n ) = Ψ ( n − as sets. We define the order onΨ ( n ) to be for t < t ′ in Ψ ( n − , we have t > t ′ if and only if π ( t min ) ≤ t < α n in Ψ ( n − and t ′ = α n ,where t min is the minimum element of D ( n − . In other words, in the ABS Ψ ( n − , we move theelement α n to the left of χ ( t min ) to construct Ψ ( n ) . We regard χ as a map on Ψ ( n ) . Then we obtainthe ABS (Ψ ( n ) , δ, χ ). We define a set D ( n ) = { t ∈ Ψ ( n ) | α n < t and χ ( t ) < α n +1 in Ψ ( n ) , with δ ( t ) = δ ( α n ) } . By hypothesis, we have D ( n ) = A ( n ) \ ˜ S ρ for all n . Here, S ρ = ( ˜ S ρ , δ ρ , π ′′ ρ ) is the ABS associatedto N ρ , where δ ρ (resp. π ′′ ρ ) denotes the restriction of δ (resp. π ′′ ) to ˜ S ρ . Hence there exists theN THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 29smallest integer a ′ such that D ( a ′ ) = ∅ . For sets Ψ ( a ′ ) , . . . , Ψ ( a ′ + n − and sets E (0) , . . . , E ( n − , letΨ ( a ′ + n ) = Ψ ( a ′ + n − as sets. The ordering of Ψ ( a ′ + n ) is given so that for t < t ′ in Ψ ( a ′ + n − , we have t > t ′ if and only if β n < t ′ ≤ π ( t max ) in Ψ ( a ′ + n − and t = β n , where t max is the maximum elementof E ( n − . In other words, to obtain Ψ ( a ′ + n ) , we move the element β n to the right of χ ( t max ). Weregard χ as a map on Ψ ( a ′ + n ) . Thus we obtain the ABS (Ψ ( a ′ + n ) , δ, χ ). We define a set E ( n ) = { t ∈ Ψ ( n ) | t < β n and β n +1 < χ ( t ) in Ψ ( a ′ + n ) , with δ ( t ) = δ ( β n ) } . By hypothesis, we have E ( n ) = B ( n ) for all n . Hence there exists the smallest integer b ′ such that E ( b ′ ) = ∅ . We obtain the admissible ABS Ψ ′ = (Ψ ( a ′ + b ′ ) , δ, χ ) which is associated to N ξ ′ with ξ ′ = ( m − f, n − g ) + ( m , n ). We immediately obtain ℓ (Ψ ′ ) = ℓ (Ψ) + 1. Hence the ABS Ψcorresponds to N − ξ ′ . Example 5.10.
Let us see an example of constructing N ξ ′ and N ρ from N − ξ . Let N ξ = N , ⊕ N , ,and we construct the full modification by 0 A and 1 B . By Example 3.9, we have N − ξ as follows: N − ξ : 1 A A (cid:9) (cid:9) A A (cid:3) (cid:3) A (cid:3) (cid:3) B A } } B B A y y A y y A y y B y y B C C B C C B v v B v v . We construct the small modification by elements 0 A and 1 A for the ABS associated to N − ξ , and weobtain N −− ξ by the following diagram which is decomposed into two cycles: N −− ξ : 1 A A (cid:9) (cid:9) A (cid:9) (cid:9) B B B A { { A { { B { { B C C B C C B w w B w w ⊕ A > > A (cid:9) (cid:9) A (cid:9) (cid:9) A (cid:9) (cid:9) . It is clear that the latter component is associated to the simple DM N , . Let us see that theformer component N is a specialization of a minimal DM . We construct the small modification byelements 1 B and 0 A , and we obtain the following diagram.1 A : : A (cid:8) (cid:8) A (cid:8) (cid:8) A (cid:8) (cid:8) A (cid:8) (cid:8) ⊕ B × @ @ B × @ @ B ◦ I I B | | B B B B B B B x x B x x . Clearly the former summand is associated to N , . For the notation of Proposition 5.9, we have D (0) = ∅ and E (0) = { B , B } . We move the element 0 B to between 1 B and 0 B . Thus we seethat the latter summand is associated to the simple DM N , . Therefore we see that this diagramcorresponds to N , ⊕ N , , and we have N −− ξ = N − (1 , , ⊕ N , .Here, for an element t of C , we give a condition of t belongs to C as follows: Proposition 5.11.
For a good exchange of 0 Ai ∈ C and 1 Bj ∈ D ′ , we have an element t of ˜ S ′ satisfying 0 Ai − < t < Bj in S − if and only if there exists a non-negative integer n such that the set π ( A ( n ) ) has the maximum element 0 Ai − .0 NOBUHIRO HIGUCHI Proof.
In the ABS ( S (0) , δ, π ), clearly there exists no element t satisfying 0 Ai − < t < Bj . Henceevery element t between 0 Ai − and 1 Bj in S − is of the form π m (0 Ai ) for a natural number m with m ≤ a . Fix a natural number n . By definitions of A ( n ) and S ( n ) , there exists an element t between0 Ai − and 1 Bj in S ( n +1) if and only if π ( A ( n ) ) has the maximum element 0 Ai − . Indeed, the element t is obtained by t = π n +1 (0 Ai ).We consider the case that C is not empty. We set an order on the set C which plays an importantrole. Notation 5.12.
Put ν = | C | . For x = 1 , . . . , ν , let i x be the natural number with m < i x ≤ h such that 0 Ai x is the element of C appearing in the x -th in the path Q . In the other words, we put C = { Ai , Ai , . . . , Ai ν } , where for elements 0 Ai x = π l x (0 Am +1 ) and 0 Ai y = π l y (0 Am +1 ) of C , we have x < y if and only if l x < l y . The elements 0 Ai x , for x = 1 , . . . , ν , of C appear in the path Q as follows. Q : 0 Am +1 → · · · → Ai → · · · → Ai → · · · → Ai ν → · · · → A m . Here, we give a characterization of the “first” element 0 Ai of C in Lemma 5.13. Lemma 5.13.
If there exists the minimum number y such that the element t = π y (0 Am +1 ) of Q satisfies 0 Am +1 < t < An in A , then t = 0 Ai . Proof.
First, let us see that for an small modification by 0 Ai and 1 Bj , there exists no non-negativeinteger n such that A ( n ) contains the element 0 Am +1 . If A ( n ) contains 0 Am +1 for a non-negativeinteger n , then π n +1 (0 Ai ) < A holds. This is a contradiction.To see that t belongs to C , for the ABS S of N ξ , we consider an small modification π by t and1 Bj with 1 Bj ∈ D ′ . Let S ( n ) and A ( n ) denote sets obtained by the small modification by t and 1 Bj .Assume that for a non-negative integer n , the set A ( n ) contains the inverse image t ′ of 1 Bj for π .In the ABS A , this t ′ is the inverse image of t . It is clear that t ′ belongs to Q . By Corollary 3.12,there exists an element t ′′ of A (0) such that π n ( t ′′ ) = t ′ . By definition of t , and since elements 1 Am and 0 An belong to P , elements τ Ax , with τ = 0 or 1, of the path Q between 0 Am +1 and t satisfies x < m or n < x . It implies that these elements do not belong to A (0) ⊂ { Am +1 , . . . , An } . Hence t ′′ belongs to P . Then the path from t ′′ to t ′ through 0 Am +1 , i.e., there exists a natural number m such that π m ( t ′′ ) = 0 Am +1 with m < a . This contradicts with the above property.We consider the case C = ∅ . Then we have the element 0 Ai of C . To apply Proposition 5.9, it isimportant to study elements between 0 Am +1 and 1 Am in S − . The following set given in Notation 5.14and the element of C given in Proposition 5.15 are useful for studying the elements between 0 Am +1 and 1 Am in S − . Notation 5.14.
For an element 0 Ai of C ′ , we often write A ( n ) i for sets A ( n ) obtained by the smallmodification by 0 Ai and 1 Bj to avoid confusion. Moreover, we often write a i for the smallest non-negative integer a satisfying A ( a ) i = ∅ . For sets {A ( n ) i } n =0 ,...,a i with 0 Ai ∈ C , we set T i = π ( A ( a i − i ) . This set consists of all elements t satisfying 0 Am +1 ≤ t < Am in S − .N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 31 Proposition 5.15.
Put i = n − γ , with γ = | T i | . Then 0 Ai belongs to C . Moreover, T i = T i holds. Proof.
Since T i x consists of some elements of { Am +1 , . . . , An } , we have | T i x | < n − m for all0 Ai x ∈ C . Hence we have m < i < n . To show that 0 Ai belongs to C , it suffices to see thatthere exists a natural number m such that A (0) i = A ( m ) i . In fact, if this statement is true, then A ( n ) i = A ( m + n ) i holds for all n with 0 ≤ n ≤ a i . Put α = π m − (0 Ai ). Note that α is the inverse imageof 0 Ai in the ABS’s ( S ( n ) , δ, π ), where π is the small modification by 0 Ai and 1 Bj . By Proposition 3.13,sets A ( m + n ) i does not contain α . Hence sets A ( n ) i do not contain the inverse image of 0 Ai for all n ,and this completes the proof. Let us see that there exists such a number m . By the definition of γ and Lemma 4.16, there exists a natural number m ′ such that A ( m ′ ) i = { Am − γ +1 , . . . , Am } . Wehave then A ( m ′ +2) i = { An − γ +1 , . . . , An } , and this set is equal to A (0) i . Therefore the number m isobtained by m = m ′ + 2. The latter part | T i | = | T i | follows from the former part.Let d be a natural number satisfying i d = n − γ . From now on, we fix notations of the non-negative integers d and γ . As can be seen in Proposition 5.16, this natural number d plays animportant role to show the former part of Proposition 5.2. Proposition 5.16.
Let 0 Ai x be an element of C . For the above natural number d , we have(1) If x ≤ d , then 0 Ai x belongs to C ,(2) If x > d , then 0 Ai x belongs to C .Thanks to this proposition, we can show the case of n > m + 1 for Proposition 5.2. Proof of Proposition 5.2 (I).
If Proposition 5.16 is true, then the equality (10) of Proposition 5.2holds for DM ’s N − ξ which are obtained by small modifications by elements 0 Ai and 1 Bj of C = C ′ \ { An } and D ′ respectively. Let us see the case of i = n . In this case, we have A (0) = ∅ ,and there exists no element t of S − satisfying 0 Ai − < t < Bj in S − . Hence we obtain desired N −− ξ = N − ξ ′ ⊕ N ρ by the case (b).Let us show some properties of sets T i , the natural numbers d and γ . These properties are usedfor showing Proposition 5.16. Proposition 5.17.
The following are true:(i) If x < y , then T i x ⊂ T i y holds;(ii) For all n with n < a i d , we have |A ( n ) i d | = γ ;(iii) For all x and n with n < a i x , we have |A ( n ) i x | ≥ γ ;(iv) T i d ( T i x holds for all x with x > d ;(v) T i x = T i d is true if and only if x ≤ d .2 NOBUHIRO HIGUCHI Proof.
Let us see (i). By assumption, there exists a natural number n such that π n (0 Ai x ) = 0 Ai y and n < a i x . We have then A ( n ) i x = { Ai y +1 , . . . , Az } , with z ≤ n . Hence A ( n ) i x ⊂ A ( m + n ) i y holds for allnon-negative integers m , and it induces the desired relation.It is clear that |A ( n ) | ≥ |A ( n +1) | for all n and all elements 0 Ai of C . By the definition of i d andProposition 5.15, we have |A (0) i d | = |A ( a − i d | = γ for a = a i d . Thus (ii) holds.By (i) and the definition of γ , we have | T i x | ≥ γ for all x . Moreover, it is clear that |A ( n ) i x | ≥ | T i x | for all n and x . Hence we see (iii).To see (iv), we fix a natural number x satisfying x > d . It suffices to see that | T i x | is greaterthan γ . Suppose that | T i x | = γ . Then there exists the minimum number u such that |A ( u ) i x | = γ .We have A ( u ) i x = { Am − γ +1 , . . . , Am } and π u (0 Ai x ) = 1 Am − γ . Then we have a sub-path of Q :0 Ai d → · · · → Ai x → · · · → Am − γ → Ah − γ → An − γ which implies that there exists a natural number m such that π m (0 Ai d ) = 0 Ai d for m < m + n . Thisis a contradiction.The statement (v) follows from Proposition 5.15, (i) and (iv).The natural number d ′ introduced in Notation 5.18 has a relation with the natural number d .We give the proof of Proposition 5.16 using this relation. Notation 5.18.
For the admissible ABS S of N ξ and the set C , we define a set L by L = { x ∈ ¯ C | For a non-negative integer n = n ( x ) , the maximum element of A ( n ) i d is 0 Ai x − } , where ¯ C = { x ∈ N | ≤ x ≤ | C |} . Let d ′ denote the maximum element of L . If L is empty, thenwe set d ′ = 0. Proposition 5.19.
For the set L and the natural number d ′ of Notation 5.18, if L is not empty,then L = { , , . . . , d ′ } . Proof.
Fix a natural number x satisfying x ≤ d ′ . We show that 0 Ai x − belongs to A ( u ) i d for a non-negative integer u . Let n be the non-negative integer satisfying that the maximum element of A ( n ) i d is 0 Ai d ′ − . Let us consider the path consisting of maximum elements of A ( m ) i d for 0 ≤ m ≤ n An → · · · → Ai d ′ − . (15)For the sub-path of Q Am +1 → A → An +1 → · · · → Ai x → · · · → Ai d ′ , since this path contains 0 Ai x , the path (15) contains 0 Ai x − . This completes the proof. Proposition 5.20.
For the natural number d ′ of Notation 5.18 and the natural number d obtainedby Proposition 5.15, we have d ′ ≤ d .N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 33 Proof.
First, let us see that there exists no non-negative integer n such that the maximum elementof A ( n ) i d is 1 Am − . Assume that the set A ( n ) i d has the maximum element 1 Am − for a non-negativeinteger n . Then this set has the minimum element 1 Am − γ , and we have π (1 Am − γ ) = 0 Ai d + m . Itimplies that the set A ( n +1) i d contains the inverse image of 0 Ai d , and this is a contradiction.Suppose d < d ′ , and we fix a natural number x satisfying d < x ≤ d ′ . Let us consider the ABS( ˜ S ′ , δ, π ) obtained by the small modification by 0 Ai d and 1 Bj ∈ D ′ . By Proposition 5.19, we obtainthe path which consists of maximum elements of A ( n ) i d for all n and T i d as follows:0 An → · · · → Ai x − → π (0 Ai x − ) → · · · → Am + γ . (16)Let us consider the sub-path of A = ( ˜ A, δ A , π A ):0 Ai x → π A (0 Ai x ) → · · · → Am + γ +1 . (17)Since the path (16) does not contain 1 Am − , the path (17) does not contain 1 Am . Hence for theABS ( ˜ S ′ , δ, π x ) obtained by the small modification by 0 Ai x and 1 Bj for S , we have a natural number n satisfying that π nx (0 Ai x ) = 0 Am + γ +1 with n < a i x . By Proposition 5.17 (iv), the set T i x contains0 Am + γ +1 . This contradicts with Proposition 3.13.We fix notation of the non-negative integer d ′ , and we give a proof of Proposition 5.16 as follows. Proof of Proposition 5.16.
For the ABS S − , let Ψ and S ρ be components of S − given by (a) or(b) of Proposition 5.2, where S ρ is the ABS associated to the DM N ρ . Let us see the case (1)of Proposition 5.16. Note that S ρ obtained by (a) of Proposition 5.2 consists of all elements of P .Recall that T i x is the set which consists of elements t satisfying that 0 Am +1 ≤ t < Am in S − . Toapply Proposition 5.9, we will show that if x ≤ d , then all elements of T i x except for 0 Am +1 belongto ˜ S ρ for the small modification by 0 Am +1 and 1 Am . By Proposition 5.17 (v), it suffices to showthat each element of Q except for 0 Am +1 does not belong to T i . In fact, this claim induces that allelements of T i belong to P and ˜ S ρ . We put two sub-paths ω and ω of Q as follows: Q : 0 Am +1 → · · · → Ai + m | {z } ω → Ai → · · · → A m | {z } ω . It follows from Proposition 3.13 that every element, which is of the form π m (0 Ai ) with m < a i ,of ω does not belong to T i . The property, which is given in Lemma 5.13, of i implies that allelements of ω except for 0 Am +1 do not belong to T i which is a subset of { Am +1 , . . . , An } .Next, let us see the case (2). We consider the small modification by 0 Ai x − and 1 Bj in S − . ByProposition 5.9, it suffices to show that if x > d , then there exists no element t between 0 Ai x − and1 Bj in S − . For the natural number x with x > d , by Proposition 5.20, clearly x > d ′ holds. To lead acontradiction, let us suppose that there exists an element between 0 Ai x − and 1 Bj in the ABS S − . ByProposition 5.11, there exists a non-negative integer v such that 0 Ai x − is the maximum element of π ( A ( v ) i x ) =: T . For sets {A ( n ) i d } n =0 ,...,a − and T i d , let Φ be the path consisting of maximum elementsof these sets: Φ : 0 An → · · · → Am + γ . m ′ to be A ( m ′ ) i x = { Am − u +1 , . . . , Am } , with u = | T | . Since m ′ isthe minimum number satisfying that |A ( m ′ ) i x | = | T | , we have m ′ < v . Put m = m ′ + 2. Then A ( m ) i x has the maximum element 0 An . We consider the path consisting of the maximum elements of sets A ( m ) i x , A ( m +1) i x , . . . , T , and we obtain the path O from 0 An to 0 Ai x − . Here, let us show that O can beregarded as a subset of Φ. It suffices to see that O does not contain 0 Am + γ . If 0 Am + γ is containedin the path O , then | T i x | ≤ γ follows from Proposition 4.17. This contradicts with the hypothesis x > d and Proposition 5.17 (iv). Hence O is a subset of Φ, and it implies that for a non-negativeinteger n , the set A ( n ) i d contains the maximum element 0 Ai x − with x > d ′ . This contradicts with thedefinition of d ′ . Here, let us see the remaining case n = m + 1 for ξ = ( m , n ) + ( m , n ), and we will give aproof of Proposition 5.2 (II). The discussion of this proof is given by the same way as the proofof Proposition 5.2 (I). As we have seen in Proposition 5.5, for every good exchange of 0 Ai and1 Bj with 1 Bj ∈ D , we have 0 Bm +1 < Bm in S − . Assume n = m + 1 for the Newton polygon ξ = ( m , n ) + ( m , n ). In this case, we have C ′ = { An } . Let us see the case D = ∅ . Notation 5.21.
For the simple DM N m ,n and its ABS B , we define sub-paths U and V of B bythe following: U : 1 Bm → Bh → Bn → · · · → Bm − n +1 ,V : 0 Bm +1 → B → Bn +1 → · · · → Bm − n . We clearly have U ⊔ V = B as sets.The above components U and V of B are useful. Concretely, as can be seen in Lemma 5.22, allelements of D belongs to the component U . Moreover, in the case (c) of Proposition 5.2, the ABScorresponding to N ρ consists of all elements of V . Lemma 5.22.
For the above notation, D ⊂ U holds. Set ι = | D | . Let j , . . . , j ι be natural numberssuch that 1 Bj x is the element of D appearing in the x -th in the path U :1 Bm → · · · → Bj → · · · → Bj → · · · → Bj ι → · · · → Bm − n +1 . We have then j = n . Proof.
For the former part, we see that there exists no non-negative integer n with n ≤ b such that π n (1 Bj ) = 1 Bm for every element 1 Bj of D by the same reasoning as Proposition 5.4. A proof is givenby the same way as Lemma 5.7.Let us see the latter part. By the former part and the hypothesis 1 ≤ j ≤ n , if 1 Bn belongs to D , then immediately we see j = n . Suppose that the set B ( n ) contains 0 Bh which is the inverseimage of 0 Ai in S ( n ) for a non-negative integer n . We have then 0 Bh < π n (1 Bj ) in S ( a + n − . Since0 Bh is the maximum element of the ABS S , this is a contradiction. Definition 5.23.
Let D (resp. D ) be the subset of D ′ consisting of 1 Bj satisfying that for ageneric specialization S − obtained by 0 Ai ∈ C ′ and 1 Bj , we have the equality (10) by the case (c)(resp. (d)) of Proposition 5.2.N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 35We give a key element of D in Proposition 5.25 to show Proposition 5.2 (II). For the abovenotation, this element characterize the sets D and D as seen in Proposition 5.26. Sets given inNotation 5.24 are used for introducing the key element and describing the proof of Proposition 5.26. Notation 5.24.
For an element 1 Bj of D ′ , we often write B ( n ) j for sets B ( n ) to avoid confusion.Moreover, we often write b j for the smallest non-negative integer b satisfying B ( b ) j = ∅ . For sets {B ( n ) j } n =0 ,...,b j , we define Z j = π ( B ( b − j ) . This set consists of all elements t satisfying 0 Bm +1 < t ≤ Bm in S − . Proposition 5.25.
Put j = 1 + µ , with µ = | Z j | . Then 1 Bj belongs to D . Let e be the naturalnumber satisfying j = j e . Then Z j = Z j e holds. Proof.
By Lemma 4.16, there exists a natural number n such that B ( n − j = { Bm +1 , . . . , Bm + µ } .We have then B ( n ) j = { B , . . . , Bµ } . This set is equal to B (0) j . A proof is given by the same way asProposition 5.15.Proposition 5.26 is shown in the same way as the proof of Proposition 5.16. The proposition isused for the proof of Proposition 5.2 (II). Proposition 5.26.
Let j x be an element of D . We have then(1) If x ≤ e , then 1 Bj x belongs to D ,(2) If x > e , then 1 Bj x belongs to D . Proof.
Let S − be the generic specialization obtained by the exchange of 0 Ai and 1 Bj in S . Forthe ABS S − , by (c) or (d) of Proposition 5.2, we obtain two components Ψ and S ρ , where S ρ isassociated with N ρ for ρ = ( f, g ). Since S ρ coincides with the component obtained from B byapplying [1, Lemma 5.6] to the adjacent 1 Bm Bm +1 or 1 Bj Bj +1 , we have gm − f n = 1. In thesame way as Proposition 5.9, we have the property: If there exists no element t of Ψ satisfying that0 Bm +1 < t < Bm (resp. 0 Ai < t < Bj +1 ) in S − , then 1 Bj belongs to D (resp. D ).First, let us show the statement (1). We have properties(i) If x < y , then Z j x ⊂ Z j y holds;(ii) For all n with n < b j e , we have |B ( n ) j e | = µ ;(iii) For all j x and all n with n < b j x , we have |B ( n ) j x | ≥ µ ;(iv) Z j e ( Z j x holds for all x with x > e ;(v) Z j x = Z j e is true if and only if x ≤ e .These properties are shown by the same way as the proof of Proposition 5.17. By (v), it suffices toconsider the case x = 1 to show the statement (1). Note that S ρ obtained by the small modification6 NOBUHIRO HIGUCHIby 0 Bm +1 and 1 Bm consists of all elements of V . There exists no element t of U satisfying that0 Bm +1 < t < Bm in S − . In fact, for the path U :1 Bm → Bh | {z } → Bj → · · · → Bm − n +1 | {z } , clearly if t = 1 Bm or t = 0 Bh , then t does not satisfy 0 Bm +1 < t < Bm in S − . By Proposition 3.17,every element of the latter sub-path, which is of the form π m (1 Bj ) with m < b , does not belong to Z j . It induces that all elements in between 0 Bm +1 and 1 Bm in S − do not belong to Ψ.Next, let us show the statement (2). We define a non-negative integer e ′ to be the maximumnumber of the set { x ∈ ¯ D | Bj x +1 is the maximum element of B ( n ) j e for n = n ( x ) } , where ¯ D = { x ∈ N | ≤ x ≤ ι } . If this set is empty, then we define e ′ = 0. By the same way as Proposition 5.20, wehave e ′ ≤ e . For these notation, a proof is obtained by the same way as the proof of Proposition 5.16(2). Proof of Proposition 5.2 (II).
By Proposition 5.26, for every element of D = D ′ \{ B } , we constructthe equality (10) by (c) or (d). Let us see the remaining case j = 1. If n >
1, then we have 0 Ai < Bj +1 and there exists no element t in between 0 Ai and 1 Bj +1 . Hence we obtain the equality (10) by (d).Suppose n = 1. In this case, we construct small modification by elements 0 Ah and 1 Bn +1 in S − , andwe obtain two components Ψ and S ρ of S −− , where ρ = (1 , S ρ = 1 B Ah .Let N be the DM associated with Ψ. We have then N = N − ξ ′ with ξ ′ = ( m − , n −
1) + ( m , n ).Hence we obtain the equality (10) by (e). In this section, we show Proposition 5.1. Theorem 1.3 follows from this proposition.
Proof of Proposition 5.1.
By Proposition 5.2, it remains to show the case λ = 1 or λ = 0. If λ = 1, then for S − obtained by a good exchange of 0 A and 1 Bj , we get ABS’s corresponding to N − ξ ′ and N ρ by (c) or (d) of Proposition 5.2. If λ = 0, then for S − obtained by a good exchange 0 Ai and 1 B , we get ABS’s corresponding to N − ξ ′ and N ρ by (a) or (b) of Proposition 5.2.Finally, we prove Theorem 1.3. Proof of Theorem 1.3.
The assertion is paraphrased as follows: For any generic specialization N − ξ of the DM N ξ with ξ = ( m , n ) + ( m , n ), there exists a Newton polygon ζ such that N ζ appearsas a specialization of N − ξ , and ζ ≺ ξ is saturated. We show this by induction on height of ξ .If the height of ξ is two (the case that the height is minimal), then ξ = (0 ,
1) + (1 , N − ξ = N , holds, whence there is nothing to prove.Assume that the height of ξ is greater than two. By Proposition 5.1, we obtain Newton polygons ξ ′ and ρ such that N −− ξ = N − ξ ′ ⊕ N ρ , where the area of the region surrounded by ξ , ξ ′ and ρ is one.By the hypothesis of induction, there exists a Newton polygon ζ ′ such that ζ ′ ≺ ξ ′ is saturated and N ζ ′ is a specialization of N − ξ ′ . If we put ζ = ζ ′ + ρ , then ζ ≺ ξ is saturated, and N ζ is a specializationof N − ξ ′ ⊕ N ρ (= N −− ξ ), and therefore is a specialization of N ξ (cf. [3, Proposition 3.5]).N THE BOUNDARY COMPONENTS OF CENTRAL STREAMS 37 Acknowledgments
I thank the referee for careful reading and helpful comments. This paper is written when the authoris a Ph.D student. I thank the supervisor professor Harashita for the constant support from theearly stage of this paper.
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