The bijection C

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

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

Cⁱ(a₁a₂. . . an) =a₁a₂. . . a_i−1a_i+₁a_i+₂. . . a_i+_ai−1a_i+_aia_i

| {z }

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

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 ∈ RCC_n we have a_i ≤ n−i and there are n−i digits to the right of a_i. Hence the action of Cⁱis at least performable.

Now let j ∈ [n]. By the definition of Cⁱ we know that C^j+1◦ · · · ◦Cⁿ will not affect the position of a_j, and not until C^jact on this word does the position of a_j move to the right with a_j positions. Moreover, those letters moving from the right of a_jto the left of a_jwill not change the relative order with a_jwhen we perform C¹◦C²◦ · · · ◦C^j−1.

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

b₁b₂. . . bn∈CC_n

and C is well-defined. 

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

D := Dⁿ◦Dⁿ⁻¹◦ · · · ◦D¹, where Dⁱis defined by

Dⁱ(a₁a₂. . . a_n) =a₁a₂. . . a_i−ai−1a_ia_i−aia_i−ai+₁. . . a_i−1

| {z }

a_i+₁a_i+₂. . . a_n. In other words, Dⁱmoves the i-th letter a_i to the left of distant a_i. 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. . . a_n∈CC_nwe have a_i ≤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 D^j−1◦ · · · ◦D¹ will not affect the position of a_j. And not until D^j applies on the word does the letter a_j move to the left with distance a_j. Moreover, those letters moves from the left of a_j to the right of a_j will not change their relative positions with a_jwhen acted by Dⁿ◦Dⁿ⁻¹◦ · · · ◦D^j+1.

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

b₁b₂. . . b_n ∈RCC_n.

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 ∈ RCC_nand suppose D◦C(τ) =b₁b₂. . . b_n. 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=idRCC_nand C◦D=idCC_n.

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= a_n.

Proof. Suppose C(a₁a2. . . an) =. . . ana_i1a_i2. . . a_ik. Note that the effect of C is to move letters to the right, and initially a_nis at the rightmost position.

Hence after the action of C, the letters a_i1, a_i2, . . ., a_ik to the right of a_nmust been moved before, which means

a_i1, a_i2, . . . , a_ik >₀ as numbers.

Now we look at D(. . . ana_i1a_i2. . . a_ik). Since the first n−k−1 move-ments only move the letters to the left of a_n, hence

D^n−(k+1)◦ · · · ◦D¹(C(τ)) =. . . a_na_i1a_i2. . . a_ik. As a₁. . . a_n∈RCC_n, a_n =0, therefore

D^n−k◦D^n−(k+1)◦ · · · ◦D¹(C(τ)) =. . . a_na_i1a_i2. . . a_ik.

From previous discussion we have a_i1 >0, hence we have

What matters is that a_i1 moves to the left to the anand the actual position of a_i1 is not important. Repeat the process on a_i2, . . ., a_ik result in Now the second lemma for the inductive step.

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

Since the action of C is to move letters to the right, hence C^m+1◦ · · · ◦Cⁿ(τ) =a₁. . . a_m[a_m+1, . . . , a_n].

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

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

For the action of C^m, we separate a_m+1, . . . , a_n into two groups, one group contains those moved from the right of am to its left after the action of C^m, 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, . . . , t⁰_β are those moved from the left to the right of a_m, and t₁, . . . , t_αare 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 a_mwill 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 D^am+α◦ · · · ◦D¹ for those to the left of a_m (namely,{s₁, . . . , s_am, t₁, . . . , t_α}) in C(τ).

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

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

{s₁, . . . , s_n−m} \ {s},

but also the letter t_j – since t_jis to the right of s_i and s is the leftmost letter among {s₁, . . ., s_n−m}, hence t_j 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, . . ., s_n−m}are to the right of a_m in τ and after the action of D◦C(τ)they will be back to the initial positions.

Hence, after the action of D^am+α◦ · · · ◦D¹letters in{s₁, . . ., s_am}will be to the right of any letters in{t₁, . . ., t_α}, as

D^am+α◦ · · · ◦D¹(C(τ)) = [t₁, . . . , tα][s₁, . . . , sam]a_m s_am+1, . . . , sn−m

t⁰₁, . . . , t⁰_β

! .

We therefore apply D^am+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, . . . , s_n−m}are to the right of a_m 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 {s_am+1, . . . , s_an−m}

will stay to the right of a_m. 

Lemma 2.3.9. Those to the left of a_m 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 s_am+1, s_am+2, . . . , s_am+j1 are still to the right of a_m. We look at the instance right before D acts on t⁰₁, which will be

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

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 CC_nis done similarly and both C and D are bijections.