The bijection C

在文檔中 A quadruple set-valued equidistribution over permutations (頁 15-21)

The bijection C is the most complicated among four. Let C : RCCn −→CCn be defined by

C :=C1◦C2◦ · · ·Cn, a composition of n mappings, where Ci is defined by

Ci(a1a2. . . an) =a1a2. . . ai1ai+1ai+2. . . ai+ai1ai+aiai

| {z }

ai+ai+1. . . an. Note the effect of Ci on the underbraced subword. In other words, Ci moves the i-th position letter ’ai’ to the right with the distance ai.

Example 2.3.1. C: RCC3 −→CC3

σ C3(σ) ≡σ0 C2(σ0) ≡σ00 C1(σ00) =C(σ)

000 000 000 000

100 100 100 010

010 010 001 001

110 110 101 011

210 210 201 012

200 200 200 002

Lemma 2.3.2. The mapping C is well defined.

Proof. For a1a2. . . an ∈ RCCn we have ai ≤ n−i and there are n−i digits to the right of ai. Hence the action of Ciis at least performable.

Now let j ∈ [n]. By the definition of Ci we know that Cj+1◦ · · · ◦Cn will not affect the position of aj, and not until Cjact on this word does the position of aj move to the right with aj positions. Moreover, those letters moving from the right of ajto the left of ajwill not change the relative order with ajwhen we perform C1◦C2◦ · · · ◦Cj1.

Since there are at least aj letters to the left of aj, hence the final position of the letter aj will be larger than the number aj. That is, in b1b2. . . bn := C(a1a2. . . an)we will have bi <i, for all i∈ [n]. Therefore,

b1b2. . . bn∈CCn

and C is well-defined. 

To prove C is a bijection we will show that there indeed exists an inverse function D. Define D : CCn −→RCCnby

D := Dn◦Dn1◦ · · · ◦D1, where Diis defined by

Di(a1a2. . . an) =a1a2. . . aiai1aiaiaiaiai+1. . . ai1

| {z }

ai+1ai+2. . . an. In other words, Dimoves the i-th letter ai to the left of distant ai. Example 2.3.3. D: CC3−→RCC3

σ D1(σ) ≡σ0 D2(σ

0) ≡σ

00 D3(σ

00) =D(σ)

000 000 000 000

010 010 100 100

001 001 001 010

011 011 101 110

012 012 102 210

002 002 002 200

Lemma 2.3.4. The mapping D is well defined.

Proof. For a1a2. . . an∈CCnwe have ai ≤i−1 and there are i−1 letters to the left of ai, hence the action of Di make senses.

Let j ∈ [n]. From the defition of Di we know Dj1◦ · · · ◦D1 will not affect the position of aj. And not until Dj applies on the word does the letter aj move to the left with distance aj. Moreover, those letters moves from the left of aj to the right of aj will not change their relative positions with ajwhen acted by Dn◦Dn1◦ · · · ◦Dj+1.

Since there are at least aj letters to the right of aj, hence the final posi-tion of the letter aj will be less than the number n−aj. In other words, in b1b2. . . bn:=D(a1a2. . . an)we have bi ≤n−i, for all i∈ [n]. Hence

b1b2. . . bn ∈RCCn.

This proves that D is well-defined. 

From the definition one may wonder that there are some symmetry be-tween C and D, and they are inverse to each other. Set τ = a1a2. . . an ∈ RCCnand suppose D◦C(τ) =b1b2. . . bn. Our goal is to prove the follow-ing.

Proposition 2.3.5. The mappings C and D are inverse to each other. Namely, D◦C=idRCCnand C◦D=idCCn.

In other words, we have

a1. . . an=b1. . . bn

and both C and D are bijections.

However this fact is never easy to prove. The bulk of this section is to prove this fact. The stategy of the proof is by Induction by combining the following two lemmas, one for the base case and the other for the inductive step.

Lemma 2.3.6. We have bn= an.

Proof. Suppose C(a1a2. . . an) =. . . anai1ai2. . . aik. Note that the effect of C is to move letters to the right, and initially anis at the rightmost position.

Hence after the action of C, the letters ai1, ai2, . . ., aik to the right of anmust been moved before, which means

ai1, ai2, . . . , aik >0 as numbers.

Now we look at D(. . . anai1ai2. . . aik). Since the first n−k−1 move-ments only move the letters to the left of an, hence

Dn−(k+1)◦ · · · ◦D1(C(τ)) =. . . anai1ai2. . . aik. As a1. . . an∈RCCn, an =0, therefore

Dnk◦Dn−(k+1)◦ · · · ◦D1(C(τ)) =. . . anai1ai2. . . aik.

From previous discussion we have ai1 >0, hence we have

What matters is that ai1 moves to the left to the anand the actual position of ai1 is not important. Repeat the process on ai2, . . ., aik result in Now the second lemma for the inductive step.

Lemma 2.3.7. Suppose we have bk = akfor k=m+1, . . . , n. Then bm =am. Proof. For convenients sake, we denote some permutation in the letters {am+1, . . . , an}by[am+1, . . . , an]. The actural order of the letters is irrelavent.

Since the action of C is to move letters to the right, hence Cm+1◦ · · · ◦Cn(τ) =a1. . . am[am+1, . . . , an].

This means that the actions of Cm+1, . . . , Cnwill not affect a1, . . . , am. Mean-while in this stage we do not care about how do Cm+1, . . . , Cnaffect am+1, . . . , an

and we can regard am+1, . . . , anas a whole.

For the action of Cm, we separate am+1, . . . , an into two groups, one group contains those moved from the right of am to its left after the action of Cm, the other contains those staying to the right of am, as

into two groups, one contains those moved from the left to the right of am

after the action, the other contains the rest, as C1◦ · · · ◦Cn(τ) =

in which t01, . . . , t0β are those moved from the left to the right of am, and t1, . . . , tαare those stay to the left.

It must be emphasized that in this step what really matters is am. For the other letters we consider the relative order with am before and after the action of C. With this there are four possibilities.

In the following we will apply D on C(τ)and use induction to explain two facts:

(i) Those to the right of amwill stay in the right of it after the action D◦C.

(ii) Those to the left of amwill stay in the left of it after the action D◦C.

These two cases will be proved respectively in the following two lemmas.

Lemma 2.3.8. Those to the right of amwill stay in the right of it after the action D◦C.

Proof. We first investigate the result after the action of Dam+α◦ · · · ◦D1 for those to the left of am (namely,{s1, . . . , sam, t1, . . . , tα}) in C(τ).

Suppose that after the action of Dam+α◦ · · · ◦D1there exists si ∈ {s1, . . . , sam} staying to the left of tj ∈ {t1, . . . , tα}. From the effect of D we know the Dn◦ · · · ◦Dam+α+1will only move am and those letters in C(τ)to the right of am. This means the relative order of si and tj is not changed. In fact, it is easy to see siwill be still staying to the left of tj in D(C(τ)).

Now we let a letter s in D(C(τ))with s ∈ {s1, . . . , snm}and s be the left most letter in any letter in{s1, . . . , snm}. We know that, in D(C(τ)), to the right of s there are not only n−m−1 letters

{s1, . . . , snm} \ {s},

but also the letter tj – since tjis to the right of si and s is the leftmost letter among {s1, . . ., snm}, hence tj is to the right of s. Hence we know that there are at least n−m letters to the right of s. This means that in D(C(τ)), s can be at one of the first m positions and this is a contradiction since the letters in {s1, . . ., snm}are to the right of am in τ and after the action of D◦C(τ)they will be back to the initial positions.

Hence, after the action of Dam+α◦ · · · ◦D1letters in{s1, . . ., sam}will be to the right of any letters in{t1, . . ., tα}, as

Dam+α◦ · · · ◦D1(C(τ)) = [t1, . . . , tα][s1, . . . , sam]am sam+1, . . . , snm

t01, . . . , t0β

! .

We therefore apply Dam+α+1on it and obtain

Hence there are at least n−m letters to the right of s, which means that in D(C(τ))the letter s can only occupy one of the first m positions, a contradiction from the fact that since letters in{s1, . . . , snm}are to the right of am in τ and after the action of D◦C they should be back to the places.

Therefore we reach the conclusion that after the action of D, letters in {sam+1, . . . , sanm}

will stay to the right of am. 

Lemma 2.3.9. Those to the left of am will stay in the left of it after the action D◦C.

From the discussion of the previous lemma we know that after the action of D the letters sam+1, sam+2, . . . , sam+j1 are still to the right of am. We look at the instance right before D acts on t01, which will be

Dam+α+1+j1◦ · · · ◦D1(C(τ)) = [t1, . . . , tα]am

Now we act by D to move t01. We need to see how the effect of C on

Proof of Lemma 2.3.7 The proof of Lemma 2.3.7 is done by combining the above two lemmas, namely

Proof of Proposition 2.3.5. By apply the above lemmas and induction on k we prove D◦C is an identity mapping on RCCn. The fact that C◦D is an identity mapping on CCnis done similarly and both C and D are bijections.



在文檔中 A quadruple set-valued equidistribution over permutations (頁 15-21)