# Proof of the main result

Now all the ingredients for the proof of the main theorem are ready.

Proof of Theorem 1.6.1 We define F : Sn →Snby F :=B◦C◦R◦A,

and for σ ∈ Sn we have sort(σ) = inv(F(σ)) (Theorem 3.1.1), Cyc(σ) = Lmap(F(σ))(Theorem 3.2.1) Lmap(σ) =Rmil(F(σ))and Lmal(σ) =Rmip(F(σ)) (Theorem 3.3.1). Hence

(sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip)

and the proof is completed. 

## Discussion and concludingremarks

It is somewhat unexpected and disappointing to us that after the comple-tion of this work we find our result can be recovered also by combining that of Chen, Gong and Guo  and Eu, Lo and Wong  together with the inverse mapping between Sn. We explain it here.

First of all, from the work of Eu, Lo and Wong  we know (sort, Cyc, Lmap, Lmal)σ= (inv, Rmil, Lmap, Lmal)φ1σ

for all σ∈Sn. For σ ∈Sn, if i∈Lmal(σ)then σj <σifor j <i. We look at σ1. As σ1(σi) = i, which means in σ1the σi-th position is i. Moreover in σ1 the position of any letter j smaller than i in σ1will be less than σi. In other words, in σ1the letters smaller than i will be to the left of i, hence i∈ Rmip(σ1).

On the other hand, suppose i ∈ Rmip(σ1), means that for any j with σ1i >σ1j we have i>j. Look at(σ1)1 =σand we know that in σ the position of i is σ1iand for any j to the left of i in σ there is σ1j < σ1iand hence i > j. In other words, in σ the letters to the left of i is smaller than i and i∈Lmal(σ).

Combine the above two paragraphs we get Lmal(σ) =Rmip(σ1). Together with Theorem 1.5.2 we obtain

Sn φ

which is exactly what we’ve done in this paper.

However, both proofs of Chen, Gong and Guo  and Eu, Lo and Wong 

are rather involved and so far it is still not clear to us that if our bijection is equivalent to the composition above. Computer experiment shows until n=5 both bijections are equivalent. We hope to clarify it in the near future.

Problem 1. Is our bijection F equivalent to φ1◦i?

A natural question arises once we have a bijection on Sn. If we apply F continuously, it is clear that each σ belongs to an orbit

σ →F(σ) →F◦F(σ) → · · · → F◦F◦ · · · ◦F(σ) →σ.

Elements in each orbit are equivalent, and the set Sncan be partitioned into equivalent classes (namely, orbits) under this equivalent relation. Hence it is natural to ask what the orbit structure is. A computer program gives the following data.

Problem 2. What is the orbit structure Snunder this equivalent relation?

n i numbers of orbits with

n i numbers of orbits with

length i in Sn length i in Sn

7

1 35

7

33 1

3 68 34 6

4 3 61 1

7 1 125 1

9 18 136 2

12 6 142 2

13 1 145 2

14 2 206 1

19 6 2894 1

24 1

Conjecture 1. There are(bnn

2c)fixed points of F on Sn.

Conjecture 2. The fixed points of F are the permutations which are both involu-tion and become a SYT which has less than two rows by RSK algorithm.

In enumerative combinatorics an important goal is the generating func-tion. For our case it seems not trivial to write down the generating function, as three of the statistics are set-valued and should be encoded by nomomi-als.

Problem 3. What can we say about the generating function F(x, y, z, w) =

### ∑

σSn

xsort(σ)yCyc(σ)zLmap(σ)wLmal(σ).

The final problem arises from many instances in the study of statistic. In the study of permutation statistics sometimes an equidistribution identity over Snholds also for some proper subset A⊂Sn.

Problem 4. Can our equidistribution result also hold for some interesting proper subset A⊂Sn?

We leave these problems for future study and also to interested readers.

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