Higher Bruhat orders and cyclic hyperplane arrangements, Topology 32

Konrad-Zuse-Zentrum für Informationstechnik Berlin

Heilbronner Str. 10, W-1000 Berlin 31

Günter M. Ziegler

Higher Bruhat Orders

and Cyclic Hyperplane Arrangements z. Zt. Institut Mittag-Leffler

Aurav?gen 17

S-18262 Djursholm

Sweden

Preprint SC 91-10 (September 1991)

Higher Bruhat Orders

and Cyclic Hyp er plane Arrangements

GüNTER M. ..ZIEGLER

Institut Mittag-Leffler

Aurav?gen 17

S-18262 Djursholm, Sweden

September 23, 1991

A b s t r a c t.

We study the higher Bruhat orders B(n, k) of Manin & Schechtman [MaS] and

- characterize them in terms of inversion sets,

- identify them with the posets ZY(C n+1'r,n+l) of uniform extensions of the alter-

nating oriented matroids C n'r for r := n—k (that is, with the extensions of a cyclic

hyperplane arrangement by a new oriented pseudoplane),

- show that B(n, k) is a lattice for k = 1 and for r < 3, but not in general,

- show that B(n, k) is ordered by inclusion of inversion sets for k — 1 and for r < 4.

However, 2?(8,3) is not ordered by inclusion. This implies that the partial order

B c (n, k) defined by inclusion of inversion sets differs from B(n, k) in general. We

show that the proper part of B c (n, k) is homotopy equivalent to S r~2. Consequently, - ￡(n, k) ~ S r~2 for J f e = 1 and for r < 4.

In contrast to this, we find that the uniform extension poset of an affine hyperplane arrangement is in general not graded and not a lattice even for r = 3, and that the proper part is not always homotopy equivalent to S r(-M'~2.

1. Introduction

The higher Bruhat orders B(n,k) were introduced by Manin & Schechtman [MaS, §2] [MaSl]. In this paper we clarify the geometric interpretation of the higher Bruhat orders (as suggested by Kapranov & Voevodsky [KaV, Sect. 4]). We use the geometric picture to analyze the main structural properties of B(n, k), including new proofs for the results of Manin & Schechtman.

We start with a review of the weak ordering of the symmetric group, see also [YaO] [Bjl] [BLSWZ, Sect. 2.3]. For this denote the set of integers {l,...,n} by [n], and the set of Jfc-subsets of [n] by ([f). We write U G U' if U, U' are finite sets with U C U' and \U'\ = \U\+1. For any collection U of finite sets, we define the partial order by single step inclusion on U by the condition that U < U' if and only if there exist sets Ui € S with U = Uo C ü! G ... G U t = U', where *'= \U'\ - \U\ is implied.

1

Definition 1.1.

(i) Let A(n, 1) denote the set of permutations of the n-element set [n].

(ii) For every permutation p = (p i,..., p…), the inversion set inv(p) := {ij : iPj} is a set of pairs, that is, a subset of (W).

(iii) Define J5(n,l) := {inv(p) : p G A(n,l)}. Every permutation is determined by its inversion set, thus A(n, 1) is in bijection to the collection B(n, 1) of inversion sets. (iv) The weak Bruhat order is "the set B(n,l), partially ordered by single step inclusion. Some main structural properties of the weak ordering are the following:

(1) B(n, 1) is a graded poset of length (!j), whose rank function is r(B) = \B\,

(2) U C (M) is an inversion set, U G B(n, 1), if and only for every triple i < j < I the

intersection U fl {ij,il,jl} is neither {il} nor {ij,jl}, [YaO, Prop. 2.2]

(3) B(n, 1) is a lattice, [YaO, Thm. 2.1]

(4) U < U' holds if and only if U C U', [YaO, Prop. 2.1]

(5) the proper part of -B(n, 1) has the homotopy type of the (r — 2)-sphere. [Bjl]

Furthermore, B(n, 1) has various geometric interpretations. For example, it is the "poset of regions" of the Coxeter arrangement A…_i, which is the arrangement of all hy-perplanes spanned by n vectors in general position in I R n-1. This also suggests far-reaching generalizations of the weak order, to the posets of regions of arbitrary arrangements. The analogues of (1), (4) and (5) are still true in this context [Edl] [EdW]. If the arrangement is simplicial, then the poset of regions is a lattice, but not in general [BEZ]. If the poset is a lattice, then an analogue of (2) holds, see [BEZ].

We will now generalize the construction of the weak orders B(n, 1) to give a definition of the higher Bruhat orders of Manin & Schechtman. The equivalence of our version of B(n, k) with the original definition is non-trivial; it will be demonstrated in Corollaries 2.3 and 4.2. Define a k-packet as the set P(I) := {J G (}y) : J G1} of all fc-subsets of a (&+l)-set I = {ii < %i < ... < ifc+i} G (fc+i)- In the lexicographic order the elements of P(I) are J\ijfc+i < I\i k <..< I\h.

Definition 1.2.

(i) A permutation p of (^) is admissible if every fc-packet P(I) occurs in it either in

lexicographic order or in reversed lexicographic order. Let A(n, k) denote the set of all admissible permutations of (^).

(ii) For each p G A(n, k) the inversion set inv(p) C (jf?^) *s the set of packets that appear in reversed lexicographic order in p.

(iii) The set B(n,k) is defined as the collection of all inversion sets B(n,k) := {inv(p) : />GA(n,fc)}.

(iv) The higher Bruhat order B(n, k) is the partial order on B(n, k) given by single step inclusion.

In this paper, we will treat the questions for higher Bruhat orders that correspond to the five structural features of the weak order listed above. In the course of our work,

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we will also show that our definition is equivalent to the original one given by Manin & Schechtman [MaS]. Specifically we prove the following results, where r := n — k.

(1) B{n, k) is a graded poset of length (￡), whose rank function is r(B) = |S|, [MaS, §2

Thm. 3b]. (Theorem 4.1(G))

(2) U C (jMj is an inversion set, U € B(n, k), if and only for every K 6 ($2) and

for {i < j < 1} C K, the intersection U D {K\l, K\j,K\i} is neither {K\j} nor {K\l, K\i}. (Theorem 4.1(B))

(3) B(n, 1) is a lattice for k = 1 and for r < 3, but not in general. (Theorem 4.4)

(4) U < U' holds if and only if U C U', provided that A r = 1 or r < 4, but not in general.

(Theorem 4.5)

This last fact shows that the (simpler) partial order B c (n, k) on the set B(n, k) defined by inclusion does not in general coincide with the partial order by single step inclusion defined by Manin & Schechtman. However, the combinatorics of B(n, k) is intimately related to that of B c(n, k), so all main results on B(n, k) have counterparts for B c(n, k), see Theorem 5.1. The partial order of B c(n, k) is easier to study, however. We prove the following result, which applies to B(n, k) whenever B c(n, k) — B(n, k):

(5) the proper part of J3c(n, k) has the homotopy type of S^r~2\ (Theorem 5.2)

The key to our development is the interpretation of B(n, k) and of B c (n, k) as "posets of oriented matroid extensions" of a cyclic configuration of n vectors in IR r by a new ele-ment. Choosing a particular vector representation, B(n, k) includes elements that corre-spond to the regions of the "adjoint" arrangement of hyperplanes spanned by the vectors, plus in general many more extensions that correspond to other extensions, realizable or not. We refer to [BLSWZ, Sect. 5.3] for the fact that the regions of the adjoint arrangement correspond to only a part of the realizable single element extensions of the correspond-ing oriented matroid. In this paper, we will treat oriented matroids as arrangements of pseudo-hyperplanes. So we get the interpretation of B(n, k) as the poset of extensions of the cyclic hyperplane arrangement X.">n~k~1 by a new pseudo-hyperplane.

This paper is organized as follows. In Section 2 we collect elementary facts about admissible orderings and show that their inversion sets are "consistent", while in Section 3 we discuss affine hyperplane arrangements and show that for "cyclic" arrangements the extensions by a new pseudohyperplane correspond to consistent sets. In Theorem 4.1, this is used for a geometric characterization of the sets A(n, k) and the higher Bruhat orders B(n, k). From this, we get in Section 4 structural information about the posets B(n, k), whose homotopy types are determined in Section 5. The geometric interpretation of B(n, k) also suggests a generalization: one can consider the poset of all extensions of any affine arrangement in general position by a new pseudo-hyperplane. This poset, however, does not retain any of the above structural features, see Section 6. Enumerative results are collected in Section 7.

3

2. Admissible Orders and Consistent Sets

We will now review the original construction of the set B(n, k) by Manin & Schechtman.

Definition 2.1. [MaS, Def. 2.2]

(i) A permutation of (^) is admissible if its restriction to each A;-packet I G (k+i) IS

either the lexicographic order or the reversed lexicographic order. A(n, k) is defined

as the set of all admissible orders on ('^).

(ii) Two permutations p, p' G A(n, k) are elementarily equivalent (p ~ p') if they differ by

an interchange of two neighbors not contained in a common packet. Let B(n, k) be

the quotient by the induced equivalence relation and A(n, k)—>? B(n, k), p i-> [p] the

quotient map.

(iii) For each p G A(n, k) the inversion set 'mv(p) is the set of packets that appear in

reversed lexicographic order in p. Here p ~ p' implies inv(p) = inv(/>'), so the inversion

set inv[/?] := inv(/>) is well-defined for [p] G B(n, k).

We will view permutations of (^') as linear orders on the set ( ^). For [p] G B(n,k)

let Q[p] be the intersection of the linear orders in [/?], that is, the partial order on ( ^ )

defined by I' < I if and only if I'

of all admissible orders r with inv(r) = 'mv(p).

Lemma 2.2. The following four sets coincide:

A\: [p], the set of linear orders of(^) equivalent to p,

A2: the linear extensions of Q[p],

A3: the admissible orders of (^') with inversion set inv(p),

A4: the linear extensions of Q'[p\.

Proof. Every r that is equivalent to p is admissible with inversion set inv(p), and thus

it is also a linear extension of Q'[p\. Now we use that any two linear extensions of a

poset Q' can be connected by a sequence of transpositions of adjacent elements that are

incomparable in Q'. Furthermore, if J, J G (^) are incomparable in Q'[p], then they are

not contained in a common fc-packet. Thus every linear extension of Q'[p] is in [p],

With this we have shown that A\ = A3 = A4. But the equality of the first and the

third set also implies Q[p] = Q'[p], that is, A\ = A3 implies A2 — A4. D

Corollary 2.3. [MaS, §2 Thm. 3d] Every [p] e B(n, k) is uniquely determined by its

inversion set inv(/>).

In particular we get from this corollary that our Definition 1.2(iii) of the set B(n, k)

is equivalent to that of Definition 2.1(ii) due to Manin & Schechtman. Our goal is

now to characterize inversion sets. For this we consider any (fc+l)-packet P(I), with

I — {h < h < ? ? ? < ü+2}5 in its lexicographic order. Thus a beginning segment is of

the form {I\ik+i,I\ik, ? ? ?,I\ij} for some j. An ending segment of P(K) is of the form

4

{A*i> A l'j-i> ? ? ? > A*i} f°r s o m e i- The subsets 0 and P(J) are considered both as begin-

ning and as ending segments of P(I). We get the following lemma, which identifies the characterizing property of inversion sets. Its converse will be proved in Theorem 4.1(B).

Lemma 2.4. Every inversion set U € B(n, k) satisfies the following equivalent conditions:

(1) U and its complement are both convex: if {ji

the intersection.of'U.with .{K\j3,K\j2,-K\ji} is neither {K\j3,K\ji} nor {K\j2},

(2) U is consistent, that is, the intersection ofU with any (k+l)-packet is a beginning or

an ending segment of it.

Proof. The condition (2) that U fl P(I) is either a beginning or a final segment of P(I)

means the following: if we consider the fc+1-packet in its lexicographic order I\ik+2 <

ü+1 < ... < ii, then there is at most one switch between elements of U and between

non-elements of U. This yields (2) ?￡=>? (1).

Now assume that p € A(n, k) is an admissible order on ( ^ ), and let {ji

for some K € Q%). Now if

K\j3 € inv(p), K\j2 i inv(/o), K\j\ € inv(p).

then this implies

K\{h,ji} >P K\{j3,j2}, K\{j2,j3}

p K\{j u j3},

which yields a contradiction. An analogous contradiction arises if we find {ii

K\j3 ￡ inv(/>), K\J2 € inv(p), K\j\ ￡ inv(p). Thus inv(p) satisfies (2). D

Let U C (^j) be a consistent set. Then the complement U := (fc+i) of U is con-

sistent as well. Define the boundary of U by dU := {/={ii< ...

U, I\ik+2 € U). Similarly, let U*{n+1} := {KU{n+l} : K € U}, and define the extension

ü C (Ij+J) of U as ü := *7*{n+l} U d*7.

The following two lemmas contain the key to an inductive treatment of consistent sets.

Their geometric significance will become clear in Section 3. In fact, both the statements

and the arguments in the proofs can be identified in Figure 1.

Lemma 2.5. Let U' be a consistent subset of ('"'), and let U" be a consistent subset

of (,W). Then U := U" U U'*{n+1} C (I^+Jl) is consistent if and only if dU' C U" and dTFcV".

Proof. Let K 6 ([?+21]). If n+1 ￡ üT, then P(K) HU = P(K) D *7" is a beginning or

ending subset of P(K), because U" is consistent.

If K = {z'i < ... < i/+i < n+1}, we let I := K\n+l and get

P(K) = {/<2f V,+1<...

5

Now if U' n P(I) is a beginning, but not an ending subset of P(I), so that i" G dU', then

U n P(K) cannot be an ending subset of P{K), so consistency if U implies I eU". From

this we get the requirement that dU' C U".

Similarly, if U'f]P(I) is an ending, but not a beginning subset of P(I), so that I € dU',

then Uf)P(K) cannot be a beginning subset of P{K), so consistency of U implies I ￡U".

From this we get the requirement that dU' Q U".

If U'CiP(I) is both a beginning and an ending segment of P(I), then either P(I) C U',

so that U n P(K) automatically is an ending segment of P(K), or P(I) n V = 0, so that

U 0 P(K) automatically is a beginning segment of P(K). From this we get that the two

conditions of the lemma are also sufficient for consistency of U. LJ

Lemma 2.6. Let U C (M) be a consistent set. Then

(i) the boundary dU ofU is a consistent subset of (}"\),

(ii) the extension UofU is a consistent subset of ( j+i )>

(iii) U is also a consistent subset of("]")?

Proof, (i) Choose K = {h < ... < i l+2} G (/"]2). We have to show that dU D P(K) is a

beginning or ending segment of P{K).

As the first case assume that üf\{zi,ij+2} ￡ J7. This implies that K\ii+2 ￡ dU. If

K\{h,i2} € ?", then this implies P(?"\*i) C C7 and thus dU n P(Jf) = 0. Otherwise we

get the existence of values s,t, with 3 < s < /+2, 1 < t < /+1 with

P(K\h) f\U = {K\{ii ,ij}:s

P(K\i l+2) C\U = {K\{i l+2, ij] : 1 < j < t}.

From this we can compute

dU D P(K) .= {{üCVJ : 1 < j < min(5 - l,i)},

which is a beginning segment of P(K).

As the second case now assume that K\{ii,ii+2} ￡ U. This implies that K\ii ￡

dU. If K\{i l+1,i l+2} <￡ U, then this implies P(K\i l+2) C ü and thus dU n P(K) =

0. Otherwise we get the existence of values s,i, with 2 < s < 1+2, 1 < t < I with

P(^V0nt7 = {Min,^}: 2 < j < a},

P(K\i l+2) nU = {K\{ii+2,ij} :t

From this we can compute

dU n P(K) = {{K\ij : max(s,i + 1) < j < 1+2},

which is a beginning segment of P(K).

(ii) Here dU' is consistent by part (i), hence we can apply Lemma 2.5 for U" = dU':

we also have dU' C U", because dU' and dU' are disjoint by definition, (iii) This is also

a special case of Lemma 2.5: for U' = 0 we get also have dU' = dW7 = 0, so U = U' is also

a consistent subset of ( ^1^). D

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3. Cyclic Arrangements

Consider any arrangement X = {Hi,..., H n} of n affine hyperplanes in general position

m JR d. Then every vertex is determined as the intersection of d hyperplanes. Associating

the vertex with the set oin—d hyperplanes that do not pass through it, we get a bijection

V = V(X) <—? (n-d)D e t w e e n the vertices of X and the (n—d)-subsets of [n]. Similarly,

every 1-dimensional line in X is the intersection of d—1 of the hyperplanes. Associating

every line with the n—d—1 hyperplanes that do not contain it, we get a bijection between

the lines of X and the (n—d+l)-sets in [n]. Furthermore, under these bijections the vertices

on a line of X correspond to the (n—rf)-sets in the corresponding (n—d+l)-set, i.e., the

vertices on a line correspond to an (n—d)-packet.

The key observation is now that if X is the cyclic arrangement of n hyperplanes in

IR n-*, then the vertices on a line correspond to a fc-packet in lexicographic order.

Definition 3.1. The cyclic arrangement X"'d is the arrangement {.Hi,... ,H n} in lR d

given by

Hi ={(x i,..., x d) € IR d : X! + tix2 + ...+ if_1x d + t{ = 0}

for 1 < i < d, with arbitrary real parameters ii < t-i < tz < ... < t n.

For every choice of the parameters ti this arrangement represents the alternating

oriented matroid c n'd+1. Here the hyperplane at infinity corresponds to the extension

n+i,d+i Q￡ Qti,d+i by a n e w element g :=n+1. Thus th^ combinatorial type of this affine

c

arrangement does not depend on the choice of the parameters tj.

L e m m a 3.2. The vertices ofX."'d correspond to (n_d) in such a way that the vertices

on an affine line correspond to the (n—d)-packets in lexicographic order (or its reverse).

Proof. There axe many ways to derive this basic fact, either by elementary linear al-

gebra, (the vertices Vj corresponding to J € (^) can be explicitly determined in terms

of Vandermonde determinants), or using simple oriented matroid tools to compute the

contractions (any contraction of C n'd is a reorientation of a cyclic oriented matroid, with

the induced linear order of the ground set), or by exploiting orthogonality resp. oriented

matroid duality. u Now consider any extension of the cyclic arrangement X",d by a new oriented hy-

perplane Hf in general position. For this, two extensions by hyperplanes Hf,Hf are

equivalent if on their negative sides they have the same set Vf = Vf> of affine vertices of

the arrangement. From Lemma 3.2 we see

Vf := {K € ( , ) '? K corresponds to a vertex on the negative side of Hf}

is a consistent set Vf C (M). The same is true for any extension of X"*r by a new oriented

pseudo-hyperplane (topologically deformed hyperplane) Hf in general position. The proper

7

Figure 1: The cyclic arrangement X^'2 with a pseudoline extension / and the corre-sponding vertex set. *

framework to study such extensions of an arrangement X by a pseudohyperplane is the theory of oriented matroids.

We will only sketch the connection, and refer to [BLSWZ] for the details. Let X — -{Hi,...,H n} be an affine arrangement in IR/*. The affine space IR d can be iden-tified with a hemisphere of S d, where the hyperplanes Hi correspond to intersections of (d— l)-subspheres of S d with the hemisphere. Assuming that a positive side has been chosen for every hyperplane, the hyperplane arrangement (resp. the corresponding sphere arrangement) represents an oriented matroid Mo of rank d+1 on the ground set [n], so the hyperplane Hi of the arrangement corresponds to the element i G [n] of the oriented matroid. By representing Mo by an affine arrangement we have distinguished the hyper-plane at infinity, which corresponds to the extension of Mo by a new element g = n+l. In this sense we say that X represents the affine oriented matroid (M,g), that is the oriented matroid Mo = M\g together with a distinguished extension of Mo by g. In particular, the cyclic arrangement X"'d represents the affine alternating oriented jnatroid (C n+1'd+l,n+l), whose structure is well understood [BLSWZ, Sect. 8.1].

The fact that the extensions of an affine arrangement by a new pseudohyperplane correspond to oriented matroid extensions is due to the "topological representation theo-rem", see [BLSWZ, Chap. 5]. Here two extensions are considered equivalent if and only

8

if they have the same set of vertices of X on their negative side, since this is equivalent

to the condition that they determine the same oriented matroid extension. Denoting by

V the vertices of X (which correspond to half of the vertices/cocircuits of the sphere rep-

resentation of .Mo), we know that every extension Mo U / of Mo is determined by its

localization, a function 07 : V —? {+,—} that indicates for every affine vertex whether it

is on the positive or on the negative side of the extension pseudo-hyperplane. See [BLSWZ,

Sect. 7.1] for details and proofs. A key technical result is Las Vergrias' characterization of

single element extensions, which in our picture can be stated as follows.

Lemma 3.3. Let V be the set of vertices of an afRne hyperplane arrangement X. A

subset Vj C V is the vertex set of an extension ofX. by a a new pseudohyperplane in

general position if and only if it contains a beginning or an ending segment of the set of

vertices on every (arbitrarily directed) line.

Definition 3.4. The uniform extension poset of X is the set of all extensions of X by

a new pseudohyperplane Hf in general position, partially ordered by single-step inclusion

of their vertex sets.

The uniform extension poset of X only depends on the affine matroid (M,g) rep-

resented by X, and will thus be denoted by U(M,g). It is the set of all uniform single

element extensions of Mo = M\g, whose partial order depends on the extension M of

Mo.

Corollary 3.5. The uniform extension poset U(C n+1'r, n+1) ofC n'r is naturally isomor-

phic to the set of all consistent subsets of(S), ordered by single-step inclusion.

Proof. This follows directly from the Lemmas 3.2 and 3.3. U To understand the geometry of U(C n+1'r,n+l) we use the partial orders of oriented

matroid programs, following the lines of [StZ, Sect. 3]. Let (M,g) be the affine matroid

of an arrangement and M = M U / an extension of M (!) corresponding to a new

pseudo-hyperplane Hf. The tripel (M,g,f) is an oriented matroid program, where Hf

is interpreted as (a level plane of) a linear objective function on the affine arrangement

{M,g), see [BLSWZ, Sect. 10.1]. The graph G f has the affine vertices of (M,g) as its

nodes, and the edges between them are the bounded edges of (M,g), directed according

to increasing /, that is, according to the direction in which their line cuts the level-plane

Hf. Assuming that M is uniform, there are no horizontal (undirected) edges. In general,

the program can be non-euclidean [EdM], so that there are directed cycles in the graph

Gf. The following non-trivial result is the technical key to our development.

Proposition 3.6. If (M,g) = (C n+1'r+1,n+l), then the graph G f is acyclic for any

program (A4 U/,g,f).

Proof. See [StZ, Prop. 4.7/Thm. 4.12]. ü

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4. Structure of Higher Bruhat Orders

With the preparations of Sections 2 and 3, we can now prove the following main theorem. Theorem 4.1. Let 1 < k < n and r := n—k.

(B) There is a naturai isomorphism of posets between

1. the higher Bruhat order B(n, k),

2. the set of all consistent-subsets of (￡"\), ordered bysingle-step inclusion,

3. the set of extensions of the cyclic arrangement X"'r_1 by a new pseudo-hyperplane

in general position, ordered by single-step inclusion of their vertex sets, and

4. the poset ZY(C n+1'r,n+l) of all uniform single element extensions ofC n'r.

(G) The poset B(n, k) is a graded poset of length (fc"1). Its rank function is r(U) = \U\.

The unique minimal element is 0 = 0, the unique maximal element is 1 = (fc+i).

(A') There is a natural bisection between

1. the set ￡(n,fc),

2. the posets Q[p], for p ￡ A(n, k), and

3. the different ways to assign directions to the 1-dimensional lines of X",r without

creating directed cycles.

(A) There is a natural bisection between

1. the set of admissible orderings A(n, k),

2. the maximal chains of the poset B(n, k—1), and

3. the different ways to sweep the arrangement X"'r by a generic pseudo-hyperplane.

We note that the geometric statements of (A3) and (B3) have precise geometric mean-

ing in the axiomatic setting of pseudoarrangements provided by oriented matroid theory,

see [BLSWZ, Chap. 5].

Theorem 4.1 also contains the main results of Manin & Schechtman: the bijection

.(A1)<->(A2) is [MaS, §2 Thm. 3c], while the part (G) is [MaS, §2 Thm. 3b]. Furthermore,

after applying oriented matroid duality (B1)?-?(B4) is a bijection between B(n, k) and the

single element liftings of C n,fe: such a bijection is stated (without a proof) by Kapranov h Voevodsky [KaV, Thm. 4.9].

Proof. We start with part (B). For this let U C (^Wj be consistent. By Corollary 2.6(h),

U is also consistent as a subset of ( ￡+j ), and thus by Lemma 3.3 it defines an extension

of X"+1'r by a new pseudo-hyperplane Hf (cf. Figure 2). We now treat H n+i as the hyperplane at infinity, with g := n+l. With this we get an oriented matroid program

(C n+i,r+i y f^n+ijy i t s^n e vertices are the vertices of X?+1'r that do not lie on

H n+i, so they correspond to the (fc+l)-subsets of [n+l] that contain n+l. The lines that

are not contained in H n+i correspond to the (&+l)-packets P(K) with n+l G K. Setting

K = {ii < i2 < ... < ik+2} and J := K\n+l, we get from Lemma 3.2 that the vertices

on such a line are given by

J = K\n+l — K\i k+l — ... — K\i2 — #\*i.

10

?!.??jraWfv &&**&&&:.

Figure 2: The cyclic arrangement Xjj'2, with g := n+1 = 6. The consistent set U = {1234,1235,1245} induces the extension by H f, directions of the lines of Xjj'2 and thus a partial order on (y).

Thus the graph Gf of the program has the vertex set V — {I € ( jj+j ) : n+l ￡ 1} = (M)*{n+1}, with directed edges (cf. Figure 2)

/ üT\i fc+i ? ... ? K\i2 ? K\h i￡JeU,

\ K\i k+1 < ... < K\i2 < K\h i￡J$U.

By Proposition 3.6, the graph is acyclic. Thus Gf defines a partial order "<" on (}f) by

I < I' :?￡=>? Gf contains a directed path from J'U{n+l} to J U{n+l},

and by construction we have

A*fc+i > ? ? ? > A?2 > A?i i f J € Ui

Aü+1 < ? ? ? < A*2 < A?i tiJ$u.

Hence every linear extension p of the partial order "<" is an admissible on (l￡J) with inv(/>) = U. With Lemma 2.4, this proves part (B).

For part (A'), this also shows that every U € B(n, k) directs the lines of X?'r in an acyclic way, and this determines a partial order Q[p] on ('￡). Finally U can be recon-structed from Q[p] as U = inv(p) for every linear extension p of Q[p\.

For part (G), we have to verify that indeed 0 < U <(^J for every consistent set U Q (i+i)'T h e r e s t i s t h e n c l e a r f r o m t h e definition of "<" by single-step inclusion. Given

11

U, we note that U = dU U U*{n+1} C ( ^t^ ) 1S consistent by Lemma 2.6(a), and thus

by (B) defines and extension Hf of X"+1,r_1. This defines a graph Gj which is acyclic

by Proposition 3.6 and thus defines a partial order ?< on (^"y as in part (A'). Any linear

extension p of this partial order is admissible, p € A(rc, k+l). By construction, U is an

order ideal of ?<, hence the linear extension

p = (Si < S2<.< 'S(k?x j) '€ A(n, k+l)

can be chosen in such a way that U is a beginning segment of p, that is, U = {Si, 52,. ? ?, S,}

for some i. However, every beginning segment {Si,52,..., S m} is consistent. Thus p

induces a maximal chain 0 = 0 < {Si} < {Si,S2} < ... < {S i,S2,..., 5/ n \} — 1

of length (fc"j) in B(k,n) that contains U.

Prom the same argument we also see (A): every admissible ordering p € A(n,k) in-

duces a maximal chain of length (￡) in B(n, k—l). According to part (G) every maximal

chain in B(n,k—1) has this form, and the linear orderings on (^) induced by maximal

chains in B(n, k—l) are clearly admissible. By (B), every maximal chain in B(n, k—l) cor-

responds to a sequence of pseudohyperplane extensions of X"'r that describes a topological

sweep, and conversely. U

The higher Bruhat order B(5,2) is drawn in Figure 3. Here every element is denoted

by the corresponding consistent vertex set of a cyclic arrangement X^'2.

We now use Theorem 4.1 to verify that our definition of the partial order on B(n, k)

coincides with the one used by Manin & Schechtman.

Corollary 4.2. B G B' holds for sets B, B' 6 B(n, k) if and only if there axe admissible

orders p, p' € A(n, k) with inv(/>) = B, inv(p') = B' and p' is obtained from p by reversing

a single k-packet P{I) whose elements appear in p in lexicographic order, with no other

?elements in between. (That is, r < r' <$=> = p7(r') in the notation of [MaS].)

Proof. The "if" part is clear. For the converse, let UCU', U'\U = {/} with / = {i x <

... < ifc+i}. Consider the line orientations of X"'r corresponding to U and to U' according

to Theorem 4.1(A'). They only differ by the reversal of one line, and both are acyclic. For

the associated partial orders Q, Q' on (^') this means that Q contains the packet P(I),

ordered lexicographically, as an interval [J\ifc+i, I\h], while Q' contains the packet P(I)^

ordered lexicographically, as an interval [I\ii, I\ik+i], arid Q and Q' differ only in the

reversal of this interval. Thus linear extensions p of Q and p' of Q' can be constructed to

satisfy the conditions of the lemma. D Now we use Theorem 4.1(B) to derive structural properties of the posets B(n, k).

Proposition 4.3. B(n, k) is isomorphic to a lower interval of B(n+1, k).

Proof. This is immediate from Lemma 2.6(iii), which shows that there is an order pre-

serving inclusion B(n, k)<—?? B(n+1, k). 0

12

13

In contrast to this, it is not clear whether there is an embedding of B(n, k) into

B(n+l,k+\) as a subposet. The case r = 2 shows that B(n,k) is not an interval of

B(n+1, fc+1) in general. The map U —> U suggested by Lemma 2.6(h) is an injection,

but not order-preserving in general.

The proofs of the following two theorems are linked. We prove, in effect, that

1) B{n, k+1) is a lattice and ordered by inclusion ==>? B(n, k) is ordered by inclusion,

2) B(n, k) is ordered by inclusion, and ri—k < 3 =4* B(n, k) is a lattice.

This can be avoided if one gives an independent proof that B(n, k) is a lattice for n—k = 3, which is possible for example by relying on geometric intuition from Figure 1.

Theorem 4.4. The poset B(n, k) is a lattice for k = 1 and for n—k < 3.

However, B(Q, 2) is not a lattice.

Proof. For k = 1 the poset B(n, 1) is the weak Bruhat order of S n, which is known to be

a lattice, see [YaO, Thm. 2.1] [BEZ].

For r < 2, the result is trivial. For r = 3, we use that B(n, k) is ordered by inclusion

by Theorem 4.5. If it is not a lattice, then by [Zie, Crit. 2] there exist six consistent sets ScSU{Ki} C T\{Lj} G T for i,j € {1,2} so that neither S?{I

K\ is the smallest and K2 is the largest set in P{K). Similarly, we get L\\JL2 =: L — [n]\h'', where L\ is the smallest and L2 is the largest set in P(L).

By symmetry, we may assume h < h!. Now if ft. = 1, then we get L2 G P{K), but

T\{L2} is consistent, so we get that K\ or K2 is not contained in T\{L2}, a contradiction.

If ft' = n, then we get K\ € P(L), but S U {-Ki} is consistent, so we get that L\ or L2 is contained in S U {-Ki}, a contradiction. Thus we have 1 < ft < ft' < n.

From K2 = [n]\{l,ft} € T\{L2) and L2 = [n]\{l,ft'} ￡ T\{L2} with K2 >lex L2 we get [n]\{l,n} i T\{L2}. From K x = [n]\{n,ft} 6 5U {K x} and L2 = [n]\{n,h'} $ So {Ki} with K x >i ex Li we get [n]\{l,n} e Si) {JCi}. But this contradicts 5U {Ki} C

T\{L2).

Now consider B(6,2) and let

5 = {123,124,356,456}, K x = 134, K

6 = 346.

Then S, Ski{K) and S'U{Z-} are consistent, while 5U{iü'i,itr6} is not consistent on

P(1346). The minimal consistent sets that contain SU{K\,KQ} are

Sn= {123,124,356,456, 134,346, 156,126, 25i} for i = 1,6.

They satisfy S?{Ki} < Sj for i,j€{l,6}, so (5U{JC1}) V (SV{K6}) does not exist. D Theorem 4.5. B(n, k) is ordered by inclusion for k = 1 and for n—k < 4.

However, i?(8,3) is not ordered by inclusion.

Proof. For k = 1 this is well known [YaO, Prop. 2.1].

14

Figure 4: The vertices and lines of B(<5,2). The vertices marked by a black dot are in

S. The vertices with a white circle are K\ and A'6.

Let Ui C U2 C (￡￡) be consistent. Then by, Theorem 4.1(B), U\ < U2 holds if

and only if there exists an admissible linear order p of (^j) of which U\ and I/2 are both

beginning segments, that is, so that Ui C C/2 are ideals of the poset Q[p]. The consistent set

U C ($,) that corresponds to Q[p] by Theorem 4.1(A') has to satisfy that f/UC/i*{n+l}

and U U J72*{"+1} are both consistent. By Lemma 2.5, this means that

8UiUdU2CU and dlhUdü^Cü. (*)

The sets dU x, dU2,?ül, dlh € B(n, Jfc+1) are consistent, with dU^dü] = 0 for i,j 6 {1,2},

so dUi C 9(7,-. By induction on r := n—k we can assume that B(n,k) is ordered by

inclusion, so dUi < dU] for i,j 6 {1,2}. For r < 4 we can assume that B(n,k+1) is a

lattice (Theorem 4.4), so a consistent set U that satisfies (*) can be chosen arbitrarily from

the interval [dUi V dU2, dUi < dlh] of B(n,k+l). Thus B(n,k) is ordered by inclusion

for all r < 4.

The smallest example we know for which U does not exist occurs in 5(8,4). This

leads to consider the consistent sets

Ui = {1234,5678}

_ fl8T\v / 1235,1245,1345,2345,1236,1246,1346,2346,1256,1

V 4 / I 4678,4578,4568,4567,3678,3578,3568,3567,3478 J

in 5(8,3) which satisfy Ui C U2. We will now give a direct proof for U x ￡ U2, which

does avoid the discussion "on the boundary". For this one first has to check that Ui and

15

V2 '-= ( 4 )\^2a f e consistent. For U\ this is obvious. For U2 one can use that the two

rows of our listing both correspond to consistent sets, which can be checked in a situation

of rank 3, since no set in the first line contains 7 or 8, while no set in the second line

contains 1 or 2. The union of both lines is consistent since no 4-packet contains sets from

both lines.

Now assume that U\ < U2. With Theorem 4.1(B) this would imply that there is a

linear order "-<" oh '(**') that orders every 4-packet either in lexicographic ("Zez") or in

reversed lexicographic ("r-Zez") order, and so that if K G Ui, K' € Ui for some i, then

K -< K'. Now we get the following sequence of implications:

1234 € U u 2347 i U x =? P(12347) lex => 1237 -; 2347 (1)

2347 € U2, 3478 i U2 =* P(23478) /ei =^ 2347 -^ 2378 (2)

(1)&(2) =? 1237-^2378 =4> P(12378) ZM =? 1237 -< 1278 (3)

2567 ￡ U u 5678 e Ui =? P(25678) r-lex =4> 2567 y- 2678 (4)

1256 i U2, 2567 e U2 =? P(12567) r-Zci ==? 1267 X 2567 (5)

(4)&(5) = =? 12671-2678 =>? P(12678) r-/ei =? 1267 >- 1278 (6)

1236 i U2, 2367 € U2 =? P(12367) r-lex =* 1237 >- 1267 (7)

where (3)&(6)&(7) yield 1237 y 1267 >- 1278 y- 1237, a contradiction. D The question whether B(n, 2) is ordered by inclusion for all n remains open. We close

this section with a list of the geometric interpretations of A(n, k) and of B{n, k) that are

available for small values of r and of k:

r = 1: A(n,n-1) = B(n,n-1) = {0,1}.

r = 2: A(n, n—2) is the set of "topological sweeps" on the cyclic line arrangement X"'2.

B(n,n—2) is the poset of consistent subsets of the affine line L[n j = P([ra]). Thus

B(n,n~2) consists of two chains of n—1 elements [MaS, §2 Lemma 7]. B(n,n—2)

can also be identified with the weak Bruhat order of the dihedral group I2(n).

r = 3: B{n,n—3) is the set of extensions of the cyclic line arrangement X￡'2 by a new pseudoline. All these extensions are in fact realizable [Ric, Thm. 8.3].

k = 1: A(n, 1) = B(n, 1) is the weak order on 5….

k = 2: .A(n, 2) is the set of maximal chains in the weak order on S n, i.e., simple allowable sequences, or arrangements of n pseudolines in "braid form" [GoP] [BLSWZ,

Chapt. 6].

P(n,2) is the set of arrangements of n+1 pseudolines that are labeled 1 to n

cyclicly at the line g := n-fl at infinity. This includes non-realizable arrangements

for n > 8. (See also [KaV, Sect. 4].) The partial order is by single-step inclu-

sion of the triples of pseudolines which determine a triangle of counter-clockwise

orientation.

The higher Bruhat orders model the set of minimal paths through a discriminantal

arrangement [MaS, §1]. By Theorem 4.1 shows that we have to choose a cyclic arrangement

for this. However, in general the poset B(n, k) contains "non-realizable" elements which

might not occur in the path space of the arrangement.

16

5. Sphericity-

There are two very natural orderings of the set B(n, k). Up to now, we have taken the

ordering by single-step inclusion as the primitive one, since it is equivalent to the ordering

defined by Manin & Schechtman, by Corollary 4.2. However, it is similarly natural (both

from a combinatorial and a geometric viewpoint), to consider the ordering of B(n, k) by-

inclusion as the appropriate generalization of the weak Bruhat order on S n. We will denote

this poset by B c (n, k). The following theorem collects its main properties.

Theorem 5.1. Let B c(n,k) := {inv(p) : p G A(n, k)} be the family of consistent sets,

ordered by inclusion.

(1) 0 = 0 is the minimal and 1 = (fc'"j) is the maximal element of B c(n, k). The length

of B c(n,k) is(^j), snd every element of B c(n,k) is contained in a maximal chain

of this length.

(2) B c(n, k) is graded for k = 1 and for r := n — k < 4, but not in general,

(2') J5c(n, k) = B(n, k) for k = 1 and for r < 4, but not in general,

(3) B c(n, k) is a lattice for k = 1 and for r < 3, but not in general.

Proof. (1) is Theorem 4.1(G). Note that (2) and (2') are equivalent restatements of The-

orem 4.5. With this, (3) is equivalent to Theorem 4.4. D The combinatorics of B c (n, k) is easier to handle than that of B(n, k). Also, in some

respects its combinatorics behaves nicer. We will now demonstrate this by computing the

homotopy type of B c (n, k).

Theorem 5.2. The proper part of the poset B c(n,k) is homotopy equivalent to an

(r — 2)-sphere:

B Q(n,k)~S r-2.

In the case k = 1 this is a result of Bj?rner [Bjl], which also follows from a theorem

of Edelman & Walker [EdW]. The geometric idea of the proof is "adjoint" to that in the

proof of [EdW]: it considers the convex hull conv(V) of the set of vertices of the affine

arrangement X(C n'r), which is a simplex, and shows that the poset ￡/(C n+1'r,n+l) is

homotopy equivalent to the face lattice of the simplex conv(V). A map between these

posets is obtained by mapping every extension to the set of vertices of conv(V) that lie

on its negative side. To see that this in fact induces a homotopy equivalence, we have to

establish several facts, which are collected in the following lemmas.

Denote by [i,j] the interval {i,i+l,... ,j} in [n], which is empty if i > j, and let

Ki := [i,i + k]. The following lemma also follows by induction from Lemma 2.6(iii).

Lemma 5.3. For all i,j e [n], the set U(i,j) := {/ G (^J : I C [i,;']} is consistent.

Proof. Let K G ($2), and note that U(iJ) H P(K) = {/ G (^J : I C K n [i,j]}.

l￡\Kn[iJ]\

P(K). In both cases U(i,j)f~l P(K) is a beginning segment.

17

Now assume \K D [i,j]\ = fc+1, with K\[i,j] = {/}. In this case we have U(i,j) =

{[i, j]}. But [i, j] is an interval, thus I is either the smallest or the largest element of K.

In the first case [ij] is the last set in the (fc+l)-packet P{K), and U(i,j) = {[i,j]} is an

ending segment. In the other case [i,j] is the first set in P{K), and U(i,j) = {[*,.?]} is a

beginning segment. U

Lemma 5.4. —The minimal-elements of -B c(n,fc)\0-are the sets {I

l

Proof. Note that {Ki} = U(i,i+k) is consistent by Lemma 5.3. By Theorem 5.1(1), the

minimal non-empty consistent subsets have exactly one element. Let K = {i\< ...

If ik+i — ii = k, then K = Ki x. Otherwise we find j ￡ K with i\ < j

{K} is not a beginning or ending segment of P(K U j), so {K} is not consistent. U

L e m m a 5.5. If U C (J j'j is consistent and K\ e U for all I e [ij], then U(i,j+k) C U.

Proof. We proceed by induction on j, the claim being trivial for j < i, where U(i,j+k) =

0, and for j = k, where U(i,j+k) = {K{}.

Let K € U(i,j+l+k)\U(i,j+k) = {I € {ll\) : j+l+k e I C [ij+i+k]} be an

element of U(i, j+l+k) that is not in U, and assume that it is selected such that the sum

of its elements is maximal.

Since K ^ [j+1, j+l+k] = Kj+i, we can find I € [j+l,j+l+k]\K. Consider the

(fc-f-l)-packet P(K U /): its smallest element (K U Z)\j+l+fc is in [i,j+fc] and hence in U

by induction. Its largest element (KUl)\ki is in U, since k\ := min K < I by construction,

hence (K U l)\k\ has a smaller sum of elements than K, and hence it is in U by the choice

of K. Thus we get a contradiction to consistency on the (A;+l)-packet P{K U /). D

Lemma 5.6. If H,ji,...,i/,j/ are such that \[i a,js] H [i*,jt]| < k for all s < t, then

U(ii,,ji) U ... U U(ii,ji) is consistent.

Proof. Suppose not, then this union is inconsistent on some (&+l)-packet P(K). However,

since each of the sets U(ii,ji) is consistent by Lemma 5.5, this requires that there are I" G

U(i s,j s)nP(K) and J € U(i tl jt)r\P{K)vnth.s < t. From this we get If)J Q'[i ai j a]0[itjt]t

hence |Z Pi?/[ < k, and 7, J cannot be contained in the same (Jb+l)-packet P(K). D

Finally we need the following "Crosscut Lemma" to establish the homotopy equiv-

alence of posets. It is a very special case of the Crosscut Theorem [Bj2, (10.8)]. It can

easily be derived from Quillen's Fiber Theorem [Bj2, (10.5)].

Lemma 5.7. Let Q be a poset with 0 and 1, and let A := min(Q) be the set of a := \A\

atoms in Q. Assume that every subset of A has a join in Q, with VA = i and Vü? < 1 for

B C A. Then Q is homotopy equivalent to the (a — 2)-sphere, Q ~ 5a~2.

Note that the joins VB exist in particular if Q is a lattice. However, we will have to

use the greater generality of the above formulation, since by Theorem 5.1(3) the posets

B c (n, k) are not in general lattices.

18

Proof of Theorem 5.2. We apply Lemma 5.7 to the bounded poset Q = B c(n,k),

whose minimal elements are {ifi}, ? ? ?, {Kr}, by Lemma 5.4. We write v(i) := {Ki}, so

A = {i>(l),..., u(r)}, and a = \A\ = r.

Every B C. A can be written uniquely as

B = {v(i) : i € [i u ji] U [i2J2] U ... U [i,,j l]} with i t+1 - j t > 2.

Now if U € B Q(n,k) = Q satisfies B C U, then I7(?i,ji+?r) U ... U U{i h ji+k) C J7

follows from consistency of C/ (Lemma 5.5). We get that U(ii,ji+k) U ... U U(ii,ji+k) is

consistent from Lemma 5.6: j3+k — i t = k — (i t — j s) < k — (it+i — js) < k — 2 for s < t

implies \[i a,js+k] n [i t Jt+k]\ = I[*t,;?+*?]! < fc - 1. Thus I7(t'i,ii+fc) U ... U U(i h ji+k)

is the join of I?, and this is 1 = U(l,n) = ￡f(l,r+&) exactly if B = {u(l),... ,u(r)} = A.

By Lemma 5.7, S c (n, fc) is homotopy equivalent to the (a — 2)-sphere. G The same proof technique cannot be used to determine the homotopy type of B(n, k)

in general: in ￡(8,3) we have atoms v(l) = {1234} and v(5) = {5678}. The sets U x =

{1234,5678} and U2 axe both upper bounds of u(l) and v(5) in ￡(8,3): for U\ this is

obvious, while for Ui it is implied by the proof technique of Theorem 4.5, since d{5678}

and ${1234} = 0. However, in Theorem 4.5 we have shown U\ ^ U2, so the atoms u(l) and

u(5) do not have a join in B(n, k). However, Theorem 5.2 implies that B(n, k) is spherical

in all cases where ￡(n, k) and B c (n, k) coincide.

Corollary 5.8. For k = 1 and for n — k < 4, the proper part of the higher Bruhat order

￡(n, k) is homotopy equivalent to an (1— 2)-sphere:

B(n, k) ~ S r~2 for k = 1 and for n - k < 4.

J t would be interesting to study the combinatorics of intervals both in B c (n, k) and in

￡(n, k). In particular, one should try to compute the M?bius function, and to determine

whether the intervals are always spherical or contractible, as they are in the case k = 1

[BjlJ-

19

十孔口琴初学者必看

初学者一开始应该怎样练习？ 初学者首先应该把口琴上的每个孔的单音吹清晰而不是急着去学压音，吹单音的方法是用发“库”的嘴型先在第四孔进行练习。 什么是压音?怎样压音？ 首先压音是十孔口琴特有的技巧，也是十孔口琴的魅力所在。压音英文也叫“bending"就是弯曲的意思，就是把口琴某一个音通过改变气流使之圆滑的下降。在十孔口琴中1到6孔可以做到吸气压音，7到10孔可以做到吹气压音。通过压音就可以使一些本身用自然吹吸奏不出的音可以很容易的演奏出来。比如第2孔的FA音和底3孔的LA音。 1，在练习之前要保证自己已经能够很好的吹，吸每一个音，如用饱满的音量吹一些很简单的曲子，如小蜜蜂之类的。 2，能分的清音的高低变化，这个很重要，可以吹吸“1” 和“7 ”这两个音来区分什么是半音，也就是小二度音程 3，发“吃” 和“去”两个中文的音，用心去感觉在发这两个音时口腔的变化方法，以及舌头的位址 4，压音技巧就是用改变吹或吸口琴的气流方向，来加大口琴吸音或吹音簧片振幅，振幅大了振动频率就小了，音高就低了。 5，我们用发“吃”的口型去吸口琴的第3孔，吸一个长音，同时不段的收紧口腔，并把舌尖顶在下鄂根部，口型是发“去”的音但是吸音！有音的变化吗？自己再多练习几边，可能不明显但应该有音降的感觉了， 好再在刚才的基础上用发“吃”的口型再去去吸口琴的第3孔，吸一个长音，同时不段的收紧口腔，并把舌尖顶在下鄂根部，口型是发“去”的音，然后在收紧，在把舌尖象下顶，把舌中部向上顶只流下很很小的缝隙在舌中部和上鄂之间，口型类似发“酷”“刻”的样子。 6，搞定了吗？压下来了吗？如没有那自己再琢磨一下。如是压下来的就是下一步了就是把音压准了 7，你发先“吃”“去”“酷”这三个字的发音有什么很大的区别吗？就是舌尖的位置越来越向下了，对这样也就是压音的方法，收紧口腔和舌中上鄂之间距离之后，通过改变舌尖的向下位置来加大压音的程度， 8，好了你家有吉他或别的乐器吗？用来做比较音高，先练：“7”这个音把就是吸第3孔的音，再用发“去”的方法降一个半音发“7b”好了在降就是“6”和“6b”了 9，练好了通知我啊我来讲下一部，没有！不可能，方法一定是对的，你才练了一天就讲自己失败了，不会把我是很笨的人，当年我在什么也不知道的情况下很吹了一个月搞定的：）我的经验一般2周可以练出了。 十孔口琴在低音部分为什么去掉4和6？ 只所以这样排列的一个主要的原因是为了演奏欧洲的一些民间音乐(因为口琴最先是在欧洲发明的,当时主要是演奏一些欧洲的音乐)比如同时吹123孔可以奏出一个C和弦(以C调口琴为例).而同时吸234孔可以奏出一个G和弦,所以都能用到"SO"这个音.以后"压音"发明了以后,才被用到演奏BLUES音乐等其他音乐中去了中去了.也许这样排列音阶的发明人没有能够预见自己的发明会被衍生出来这么多的音符和音乐，估计他要是活着也会被惊讶了。（shawlee著） 十孔口琴共有几个调子？C调口琴比D调口琴音低吗？ 十孔口琴一共有12个调子,每个调子只是在音高上不一样,其他音阶排列都是一样的,由于音高不一样,不同调子压音的感觉也是不一样的.在十孔口琴12个调子中G调口琴音最低.从低往高排列依次是:G #G A bB B C #C D bE F #F 最高的调子是#F调．以上是１２个大调口琴的音高排列．在有些品牌的型号口琴中还有几个特殊调子，说是特殊实际上就是在调子上有加了一些．在低调子中加了一个比Ｇ还要低的调子Ｆ调一般都叫它LOW F调，甚至还有比LOW F还要低的LOW E ,LOW D 等．在高调子中加上一个比#F还要高的调子Ｇ，一般都叫它H G. 例如TOMBO的FOLK BLUES就有１４个调子，１２个大调外加一个LOW F 和一个H G．通过以上介绍Ｃ调口琴在音高上显然比Ｄ调口琴低。

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目录 第一部分概况 (1) 1、项目名称 (1) 2、项目宗旨 (1) 3、经营范围 (1) 4、项目投资 (1) 5、项目概况 (1) 第二部分产品和服务 (1) 1、初期服务 (1) 2、中后期服务 (1) 第三部分市场 (2) 1、客户分析 (2) 2、需求分析 (2) 第四部分竞争 (2) 第五部分战略规划 (2) 1、宣传工作 (2) 2、销售方式 (3) 3、市场营销计划 (3) 第六部分人员及组织结构 (4) 第七部分 SWOT分析 (4) 第八部分财务规划 (4) 1、财务支出 (5) 2、月、年收入 (5) 第九部分风险与风险管 (5) 第十部分未来规划 (6)

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2020年江苏注册会计师报名时间：4月

2020年xx注册会计师报名时间：4月 5-28日报名人员应当于2017年4月5日至4月28日(网报系统开放时间为每天8:00—20:00)，点击进入网报系统进行注册，按照报名指引如实填写相关信息。首次报名人员还应在网报系统中上传符合要求的本人最近1年1寸免冠证件照片。非首次报名人员的相关信息如发生变动，应当重新填写。 无法上传照片的报名人员可在报名期间持有效身份证原件到报名所在市注册会计师协会(简称市级注协)指定地点(详见网报系统中有关各市报名信息)进行现场采集。 符合综合阶段考试报名条件，但不能进行报名的人员，可向江苏省注协查询办理。 资格审核： 1.首次报名人员填报的毕业证书编号，原则上由网报系统链接中国高等教育学生信息网进行认证。持国外学历的报名人员填报的教育部留学服务中心出具的学历认证书编号，由中注协、省注协提交教育部留学服务中心进行认证。 应届毕业生报名人员应当于7月20日至31日(网报系统开放时间为每天8:00—20:00)期间，登录网报系统补录本人的毕业证书编号或学历认证书编号，并由中注协、省注协提交相关单位进行认证。 未在规定时间内补录毕业证书编号或学历认证书编号的，不符合报名条件，报名资格不予通过。 2.持非居民身份证(如军官证、护照)，或者以军校毕业学历、会计或者相关专业中级以上技术职称作为报名条件的首次报名人员，或者网报系统无法认证学历的首次报名人员，应当在报名期间携带本人签名的报名信息表、有效身份证件、毕业证书或职称证书等相关材料，到报名所在市注协指定地点进行审核。 3.未通过资格审核的报名人员，不符合报名条件，报名资格不予通过。 交费：

1.完成报名资格审核的报名人员，以及无需进行资格审核的报名人员、应届毕业生报名人员在报名信息填写完成后，通过网上支付交纳考试报名费。交费完成视为报名程序完成。 2.报名费标准按照《江苏省物价局关于调整注册会计师考试考务费标准的复函》(苏价费函〔2013〕128号)规定，考试报名费标准为每科次92元。报名发票在报名所在市注协领取，领取时间到2017年12月31日截止。 3.交费截止时间为4月28日20:00。交费手续完成后，所报考科目及相关信息不能更改，报名费不予退还。 4.报名人员(不含应届毕业生)完成交费手续后，可在网报系统查询个人报名状态;应届毕业生报名人员可在8月15日后登录网报系统查询个人报名状态。报名资格未通过或未交费的报名人员，不能下载打印准考证和参加考试。

大学生创新创业训练计划项目申报书

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填写须知 一、“大学生创新创业训练计划”创业项目申报书请按顺序逐项填写，填写内容必须实事求是，表达明确严谨。空间不够，自行加页。空缺项要填“无”。 二、格式要求：表格中的字体小四号仿宋体，倍行距； A4纸双面打印，于左侧装订成册；“签名”栏由相关人员用黑色钢笔或水笔签名。 三、申请参加“大学生创新创业训练计划”创业项目团队的人数含负责人在内不得超过5人。

目录 一、项目申请人及参加人情况....................................错误!未定义书签。 二、项目方案..................................................错误!未定义书签。 （一）立项依据............................................错误!未定义书签。 1、项目提出背景：.................................错误!未定义书签。 2、行业分析：.....................................错误!未定义书签。 3、市场调研：.....................................错误!未定义书签。 4、可行性分析：...................................错误!未定义书签。 5、理论与实践意义：...............................错误!未定义书签。 （二）项目内容............................................错误!未定义书签。 1、企业执行状况简介...............................错误!未定义书签。 2、市场分析.......................................错误!未定义书签。 3、营销策略.......................................错误!未定义书签。 4、人员与组织结构.................................错误!未定义书签。 5、财务分析.......................................错误!未定义书签。 6、利润预测.......................................错误!未定义书签。 7、风险分析及预测.................................错误!未定义书签。 8、市场远景.......................................错误!未定义书签。 （三）项目实施进度安排....................................错误!未定义书签。 （四）成果提供形式........................................错误!未定义书签。 （五）项目验收要点........................................错误!未定义书签。 （六）其它支撑项目的证明材料.............................错误!未定义书签。 三、项目经费预算..............................................错误!未定义书签。 四、指导教师简况..............................................错误!未定义书签。 五、指导教师对该项目的评价....................................错误!未定义书签。 六、推荐系部审核意见..........................................错误!未定义书签。 七、学校领导小组意见..........................................错误!未定义书签。 八、学校评审意见..............................................错误!未定义书签。

注册会计师(CPA)报名条件具有下列条件之一的中国公民,可报名参加考试

注册会计师（CPA）报名条件具有下列条件之一的中国公民，可报名参加考试： 1、高等专科以上学校毕业的学历； 2、会计或者相关专业（相关专业是指审计、统计、经济。下同）中级以上专业技术职称。 注册会计师（CPA）报名时间 注册会计师全国统一考试报名时间一般在每年的3月中旬至4月中旬（2006年是从4月1日至4月30日，原则上是不少于20个工作日），具体时间由各地方考试委员会确定，一般应不少于20个工作日。转贴于51论文网https://www.doczj.com/doc/a28700203.html, 最新：注册会计师（CPA）考试科目（一）原制度考试科目：会计、审计、财务成本管理、经济法、税法。51-论文-网-欢迎您 （二）新制度考试科目：会计、审计、财务成本管理、公司战略与风险管理、经济法、税法。51-论文-网-欢迎您 注：新制度考试划分为专业阶段考试和综合阶段考试。考生在通过专业阶段考试的全部科目后，才能参加综合阶段考试。51-论文-网-欢迎您 专业阶段考试设会计、审计、财务成本管理、公司战略与风险管理、经济法、税法6个科目；综合阶段考试设职业能力综合测试1个科目。51-论文-网-欢迎您 每科目考试的具体时间，在各年度财政部考委会发布的《注册会计师全国统一考试报名简章》中明确。51-论文-网-欢迎您 （三）考试范围：由财政部注册会计师考试委员会在发布的《注册会计师全国统一考试大纲》中确定。 新考试制度下备考建议、科目分析与报考建议注册会计师资格考试 文章作者 100test 发表时间 2009:02:04 17:09:09 来源 https://www.doczj.com/doc/a28700203.html,百考试题网 一、新考试制度下的报考建议 整体计划安排：新考试制度下一共六门课程，大体建议以每年三门的速度报考，这样两三年的时间大体可以通过专业阶段考试，两三年的时间也是考虑到了知识的更新问题，因为要应对综合阶段考试，所以考试周期最好不要拉的太长，所以建议考生认真规划自己的考试计划，每年三门基本上符合大多数人的能力和要求。 科目联系分析：会计、审计和税法是联系最为紧密的科目了，因为审计中有关内部控制的内容分出去了，那么以后考试的时候审计和会计的联系将更加强一些；税法无疑和会计相关度很大，会计中涉税会计处理也是考察的重点和难点；财管和风险管理科目无疑联系也是最为紧密的了，因为风险管理主要是从财管中分离出去的，两门课程建议一起报考比较好。另外风险管理能够帮助审计的学习，因为审计中有很多内部控制的内容，但是教材中介绍的都比较浅显，风险管理这个新科目解决了审计学习的难点和切实需要，两门也比较适合一起报考；经济法仍然是比较超然独立的科目，大家都和他没有什么关系。 科目报考建议：09年建议财管和风险管理一起报考，因为现在复习08年财管等于同时复习两门课程，何乐不为呢？而且又是政策变动第一年，财管内容少了，风险管理还不健全，所以考试难度不会很大，加上改革时候政策可能放水，那么这个机会肯定是要把握的。

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项目说明 1、深刻解读文件。 1）教育部《关于做好“本科教学工程”国家级大学生创新创业训练计划实施工作的通知》（教高函…2012?5号） 2）《教育部财政部关于实施高等学校本科教学质量与教学改革工程的意见》（教高…2007?1号） 3）《教育部关于进一步深化本科教学改革全面提高教学质量的若干意见》（教高…2007?2号）等 是本科教学质量工程的体现，是改革成效的检阅。 2、分级实施：校级、省级、国家级三级 数量：今年，国家、省级20、校级40 3、内容：三类 创新训练项目是本科生个人或团队，在导师指导下，自主完成创新性研究项目设计、研究条件准备和项目实施、研究报告撰写、成果（学术）交流等工作。-科技处创业训练项目是本科生团队，在导师指导下，团队中每个学生在项目实施过程中扮演一个或多个具体的角色，通过编制商业计划书、开展可行性研究、模拟企业运行、参加企业实践、撰写创业报告等工作。团委 创业实践项目是学生团队，在学校导师和企业导师共同指导下，采用前期创新训练项目（或创新性实验）的成果，提出一项具有市场前景的创新性产品或者服务，以此为基础开展创业实践活动。 4、前期工作： 安康学院“大学生创新创业训练计划”实施方案、安康学院大学生创新创业训练计划项目管理办法 5、实施 1）原则：“立足兴趣、统一组织、鼓励创新、强化过程、注重实效” 2）以项目形式立项实施，实行项目负责人制。项目负责人为在校全日制普通本科生选好项目是关键。 ●立项原则。坚持可行性、创新性原则、索性原则，结合学校特色，依托学科优势，产学研结合，开展立项研究与实践。同时要注重理论联系实际和实践课题实效，注重学生实践能力和创新、创业精神的培养。 ●选题要求：申报的项目应以解决科学、技术、生产、生活和社会领域的具体问题为出发点，以知识、技术和方法创新为主要目的，以实验为主要手段，目标明确、思想新颖、技术先进、创新性强、立论根据充足、方法合理、具有创新性和探索性，并且具备项目实施条件。 ●项目范围：主要包括发明、创作、设计等项目；应用性、创新性研究项目；教师科研课题中的子项目；社会调研项目；其他有价值的研究与实践项目。 3）项目周期：创新训练项目实施期限一般为1年，创业训练项目一般不超过2年，创业实践项目一般不超过3年，所有项目均须在项目负责人毕业离校前完成。 4）项目说明：围绕：目的意义、研究内容和拟解决的关键问题、基础条件、实施方案、预期成果、经费预算等开展论述。 5）申报程序：项目征集遴选（来源：学生、教师、下达）（简表）——院系筛选——推荐——学校评审——意见反馈下达——院系组织申报（申请书、汇总表）——学校汇总上报

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大学生创新创业训练计划 项目结项验收总结报告 项目类别国家级（）校级（）项目类别创新项目（）创业项目（）项目编号 CX12042 项目名称农村小学家校信息交流现状调查和实践对策 陕西师范大学大学教务处

二Ｏ一三年 总结报告参照提纲及要求： 一、题目 二、关键字 三、项目概述 1.项目成员基本情况（人数、院系、专业、年级）、指导教师 基本情况（职称、研究方向） 2.项目的研究背景、目的与意义 3.项目实施过程的人员工作分配和完成情况 4.项目实施过程收获和体会 四、项目预期成果完成情况和创新点 五、项目总结 六、附录（参考文献目录等） 用A4纸双面打印（宋体，小4号字，倍行距，首行缩进2字符，段前段后0行，默认页边距，页码居中）。

科研中成长，实践里圆梦 ——我们的科研小总结 关键词：学生科研教师梦团队协作成果满意 项目概述 家校合作，是一个农村小学教育的重大问题。如何更好地加强学校和家庭的合作已被教育界所重视,并成为农村学校改革的重要内容，也是当今农村教育发展的一个趋势。作为一名教育学本科生我与本班同学们积极关注农村小学家校合作的问题，思考如何拉近家教之间的距离，为此我们努力搜集资料，开展研究，确立了“农村小学家校信息交流现状调查和实践对策”为研究专题项目成员基本情况:我们团队人员共有4位同学，均为2010级教育学本 科生。在项目确立与进行中指导老师张正峰副教授（研究方向：高等教育、 教育基本理论）倾力支持，保证我们学生科研项目顺利实施与完成。 项目的研究背景、目的与意义； 国内研究理论体系不断完整，研究家校合作论文不断增长，家校合作模式不断转新，并且大多数研究集中在探讨家校合作的模式和存在问题。 国外研究日本PTA(Parent Teaehe：ASsoeiation),即父母与教师联合会。PTA 在致力于沟通学校与家庭、社区的联系和创造一个有利于青少年成长的环境方面发挥了巨大的作用,成为中小学教育中不可忽视的重要教育力量。日本PTA组织的许多成功经验为我国开展家校合作提供了有益的启示。形质兼备,寻求法律与制度保障，将网络服务引入家校合作，拆除学校壁垒,使学校教育教学活动透明化。日本一些学校实行教学参观活动,随时允许任何人(不只限于家长,社区里的任何人都可以)来校参观教学活动。美国“家长参与教育”（Parental Involvement in Education）逐渐成为美国教育重要形式为大幅度提高学生的学业成绩，美国的很多中小学与学生家长之间所进行的一种双向交流活动。它主要包括：家长为子女主动参与学校改革和学校为学生主动培训家长两个方面。家长为子女主动参与学校改革家长为子女主动参与学校。 项目研究意义：

注册会计师考试时间及科目

注册会计师考试时间及科目 2017注册会计师考试时间及科目 2017年注册会计师考试时间什么时候呢，考试科目有哪些呢?大家不用着急，现在官方暂时还没有公布，大家可以参考一下去年的情况。 考试时间： 综合阶段考试： 2016年8月27日 08：30—12：00职业能力综合测试(试卷一) 14：00—17：30职业能力综合测试(试卷二) 专业阶段考试： 2016年10月15日 08：30—11：00审计 13：00—15：30财务成本管理 17：30—19：30经济法 2016年10月16日 08：30—11：30会计 13：30—15：30公司战略与风险管理 17：30—19：30税法 考试科目和考试范围

专业阶段考试科目：会计、审计、财务成本管理、公司战略与风险管理、经济法、税法。 专业阶段考试报名人员可以同时报考6个科目，也可以选择报考部分科目。 综合阶段考试科目：职业能力综合测试(试卷一、试卷二)。考试范围：《注册会计师全国统一考试大纲——专业阶段考试(2016年)》和《注册会计师全国统一考试大纲——综合阶段考试(2016年)》确 定了考试范围。 报名条件： 专业阶段考试报名条件： 同时符合下列条件的.中国公民，可以申请参加注册会计师全国 统一考试专业阶段考试： 1、具有完全民事行为能力; 2、具有高等专科以上学校毕业学历，或者具有会计或者相关专业中级以上技术职称。 综合阶段考试报名条件： 同时符合下列条件的中国公民，可以申请参加注册会计师全国统一考试综合阶段考试： 1、具有完全民事行为能力; 2、已取得注册会计师全国统一考试 专业阶段考试合格证。 有下列情形之一的人员，不得报名参加注册会计师全国统一考试： 1、因被吊销注册会计师证书，自处罚决定之日起至申请报名之 日止不满5年者;2、以前年度参加注册会计师全国统一考试因违规 而受到停考处理期限未满者。

大学生创新创业训练计划-项目申报书(仅供参考)

项目编号： 中山大学南方学院“大学生创新创业训练计划” 项目申报书 系经济学与商务管理系 项目名称说不出的挂链（念） 所属一级学科名称艺术设计与高分子化学工程 所属二级学科名称工艺饰品 项目类型□创新训练项目□创业训练项目 创业实践项目 负责人 指导教师马立新 中山大学南方学院教务部 二〇一二年四月

填写须知 一、项目分类说明： 1.创新训练项目是本科生个人或团队，在导师指导下，自主完成创新性研究项目设计、研究条件准备和项目实施、研究报告撰写、成果（学术）交流等工作。 2. 创业训练项目是本科生团队，在导师指导下，团队中每个学生在项目实施过程中扮演一个或多个具体的角色，通过编制商业计划书、开展可行性研究、模拟企业运行、参加企业实践、撰写创业报告等工作。 3. 创业实践项目是学生团队，在学校导师和企业导师共同指导下，采用前期创新训练项目（或创新性实验）的成果，提出一项具有市场前景的创新性产品或者服务，以此为基础开展创业实践活动。申报该类项目需额外提交企业导师合作指导协议书作为附件。 二、申报书请按顺序逐项填写，填写内容必须实事求是，表达明确严谨。空缺项填“无”。 三、申请参加大学生创新创业训练计划项目团队的人数含负责人在内不得超过5人。 四、填写时可以改变字体大小等，但要确保表格的样式没有被改变；填写完后用A4纸张双面打印，不得随意涂改。 七、申报过程有不明事宜，请与教务部联系和咨询。

项目名称说不出的挂念 项目类型（）创新创业项目（）创业训练项目（√）创业实践项目 项目实施时间起始时间：2012年7 月完成时间：2013年12 月 申请人或申请团队 姓名年级学校 所在院系 /专业 联系电话E-mail 主持人 中山大学 南方学院 经济学与商 务管理系 中山大学 南方学院 经济学与商 务管理系 成员 中山大学 南方学院 经济学与商 务管理系 中山大学 南方学院 经济学与商 务管理系 中山大学 南方学院 经济学与商 务管理系 指导教师 姓名马立新研究方向投资与金融管理年龄45 行政职务/专业技术 职务 教师主要成果 文章发表：中国经济时报“旅游经济调整和投资重点”2000年10月25

入门口琴BD谱

Ode to Joy （欢乐颂） 5B 5B 5D 6B 6B 5D 5B 4D 4B 4B 4D 5B 5B 4D 4D 5B 5B 5D 6B 6B 5D 5B 4D 4B 4B 4D 5B 4D 4B 4B 4D 4D 5B 4B 4D 5B 5D 5B 4B 4D 5B 5D 5B 4D 4B 4D 3B 5B 5B 5D 6B 6B 5D 5B 4D 4B 4B 4D 5B 4D 4B 4B 雪绒花 5B 6B 8D 7B 6B 5D 5B 5B 5B 5D 6B 6D 6B 5B 6B 8D 7B 6B 5D 5B 6B 6B 6D 7D 7B 7B 8D 6B 7D 6D 6B 5B 6B 7B 6D 7B 8D 7B 7D 6B 5B 6B 8D 7B 6B 5D 5B 6B 6B 6D 7D 7B 7B 送别 6B 5B 6B 7B 6D 7B 6B 6B 4B 4D 5B 4D 4B 4D 6B 5B 6B 7B 7D 6D 7B 6B 6B 4D 5B 5D 3D 4B 6D 7B 7B 7D 6D 7D 7B 6D 7D 7B 6D 6D 6B 5B 4B 4D 6B 5B 6B 7B 7D 6D 7B 6B 6B 4D 5B 5D 3D 4B 虫儿飞 8B 8B 8B 9D 9B 8B 8D 8D 7B 7B 7B 8D 8B 8B 7D 7D 6D 8B 8D 6D 8B 8D 6D 8B 8D 7B 7B 8B 8D 9B 9D 8B 8D 9B 9D 8B 9D 9B 8B 8D 7B 6D 8B 8D 7B 6B 8D 7B 9D 8B 9D 8B 7B 9D 8B 9D 8B 7B 8D 7B

注册会计师报考条件

注册会计师报考条件 考试资格 具有高等专科及以上学校毕业的学历、或者具有会计或者相关专业中级以上技术职称的中国公民，可以申请参加注册会计师全国统一考试；具有会计或者相关专业高级技术职称的人员，可以免予部分科目的考试。 报名条件 （一）同时符合下列条件的中国公民，可以申请参加注册会计师全国统一考试专业阶段考试： 1. 具有完全民事行为能力； 2.具有高等专科以上学校毕业学历，或者具有会计或者相关专业中级以上技术职称。 （二）同时符合下列条件的中国公民，可以申请参加注册会计师全国统一考试综合阶段考试： 1. 具有完全民事行为能力； 2. 已取得财政部注册会计师考试委员会（简称财政部考委会）颁发的注册会计师全国统一考试专业阶段考试合格证并且在有效期内。 （三）有下列情形之一的人员，不得报名参加注册会计师全国统一考试： 1. 因被吊销注册会计师证书，自处罚决定之日起至申请报名之日止不满5年者； 2. 以前年度参加注册会计师全国统一考试因违规而受到停考处理期限未满者。 [1] 免试条件 具有会计或者相关专业高级技术职称的人员（包括学校及科研单位中具有会计或者相关专业副教授、副研究员以上职称者），可以申请免予专业阶段考试1个专长科目的考试。 免试年度为申请人提交申请的年度，申请人应当于本年度考试开始前提交免试申请。 [3] 报名方式 注册会计师全国统一考试报名通过“注册会计师全国统一考试网上报名系统”进行。报名分为网上预报名、资格审核、交费确认三个步骤. 登录中国注册会计师协会网站，点击“注册会计师全国统一考试网上报名”链接，进入“注册会计师全国统一考试网上报名系统”进行网上预报名。 首次报名参加专业阶段考试的报名人员，须持预报名信息表、身份证件原件及复印件、学历证书或中级以上职称证书原件及复印件，到报名地现场办理报名资格审核。 持有国外学历证书的报名人员，还应当提供学历证书原件以及经公证机关公证的该学历已获得国家教育部认可的中文证明。