The random (n-k)-cycle to transpositions walk on the symmetric group
aa r X i v : . [ m a t h . P R ] M a r THE RANDOM ( n-k ) -CYCLE TO TRANSPOSITIONSWALK ON THE SYMMETRIC GROUP ALPEREN Y. ¨OZDEMIR
Abstract.
We study the rate of convergence of the Markov chain on S n which starts with a random ( n − k )-cycle for a fixed k ≥
1, followed byrandom transpositions. The convergence to the stationary distributionturns out to be of order n . We show that after cn + ln k n steps for c >
0, the law of the Markov chain is close to the uniform distribution.The character of the defining representation is used as test function toobtain a lower bound for the total variation distance. We identify theasymptotic distribution of the test function given the law of the Markovchain for the ( n − Acknowledgement
The author would like to thank Jason Fulman for sug-gesting the problem and his most valuable comments.1.
Introduction
Random walks on symmetric group are widely associated with card shuf-fling problem. For an extensive survey on random walks defined on finitegroups, see [7] or [18]. In their influential work [9], Diaconis and Shahsha-hani study the random transposition walk on symmetric group by usingrepresentation theory. They demonstrate that the chain exhibits a cutoff at n ln n for the total variation distance. In recent years, more probabilisticapproaches are developed. Berestycki et al. [3] prove a cutoff at nk ln n stepsfor k − cycle walk by analyzing the evolution of the cycle distribution. Apath coupling argument is shown to be applicable for random transpositionwalk to obtain a mixing time of order n ln n in [6].We study a variation of the random transposition walk, following thetechniques developed in [9], by taking the initial permutation to be a cyclewith a fixed number of fixed points instead of a transposition. The upperbound for the ( n − − cycle to transpositions case is given in [10] along withboth an upper and a lower bound for n -cycle to transpositions walk. Thelower bound is left an open problem. Mathematics Subject Classification.
Key words and phrases.
Markov chain, Convergence rate, Symmetric Group, Definingrepresentation, Asymptotic distribution, Murnaghan-Nakayama Rule.
The chain starts at the identity. At the initial step, an ( n − k )-cycle isuniformly selected for a fixed k ≥ . From then on, the transition is select-ing a transposition uniformly and multiplying it by the permutation of thecurrent state to obtain another permutation for the next state. Observethat the chain alternates between A n and S n \ A n . Therefore the limitingdistribution is not the uniform distribution on S n unlike the random trans-position walk in [9], for which they define a holding probability that makesthe random walk lazy. We define the transition probabilities as in [10], andallow the limiting distribution to be different for even steps and odd steps.The analysis is technically the same and the limiting distribution is uniformon the corresponding coset. U t will stand for the uniform distribution onthe coset ( A n or S n \ A n ) where the chain is on, at step t .To be more precise on the stationary distribution, we identify it withrespect to the length of the initial cycle, n − k. We call two integers to havethe same parity if they are both odd or both even, otherwise we call themto have different parity. Denoting the uniform distribution over a set S by U ( S ) , we have U t = ( U ( A n ) if t and n − k have the same parity, U ( S n \ A n ) if t and n − k have different parity.Let the probability assigned to σ ∈ S n be µ t ( σ ) after t steps. The totalvariation distance between µ t and the uniform distribution U t is(1) k µ t − U t k T V = 12 X σ ∈ S n | µ t ( σ ) − U t ( σ ) | = max S ⊆ S n | µ t ( S ) − U t ( S ) | . We are interested in finding upper and lower bounds for the probabilitymetric defined above and identify the convergence rate. For the asymptoticconvergence, we use the notation f ( n ) ∼ g ( n ) , meaning lim n →∞ f ( n ) g ( n ) = 1 , where n is the number of elements that S n is defined over.The organization of the paper is as follows. In Section 2 we summarizethe group representation theory techniques that are used in the rest of thepaper. In Section 3, we focus on the ( n − Theorem 1.1. As n → ∞ , for any c > , after a random ( n − -cycle and cn random transpositions, − e e − c + e e − c − c e − o (1) ≤ k µ cn +1 − U cn +1 k T V ≤ e − c √ − e − c + o (1) . HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP In Section 4, a lower bound on the total variation distance for k ≥ cn + ln k n steps is obtained. The first two moments of the character of thedefining representation are needed to bound the distance to the stationarydistribution. Then we derive estimates for the differences of characters andcombine them with the previous results in literature to find the upper boundfor the given convergence time. This leads to the following theorem. Theorem 1.2.
Let k ≥ be a fixed number. As n → ∞ , for any c > , after a random ( n − k ) -cycle and random t = cn + n ln k transpositions, e − c − o (1) ≤ k µ t +1 − U t +1 k T V ≤ r e − e − c + o (1) . Representation theory techniques
We summarize the techniques that are employed in the following sections.The tools developed in [9] to study the random transposition walk are usedto obtain lower and upper bounds in our case. To start with the connectionto the group representation theory, first consider the
Fourier transform ofthe measure µ defined on a finite group G , evaluated at any representation λ of G ,(2) b µ ( λ ) = X g ∈ G µ ( g ) λ ( g )The inverse transform is expressed as µ ( g ) = X λ irred. d λ tr ( λ ( g − ) b µ ( λ ))where the sum is over all irreducible representations of G and d λ standsfor the dimension of the representation λ, which is the number of standardYoung tableaux of shape λ. The inverse transform leads to Plancherel’sformula stated below.(3) X g ∈ G | µ ( g ) | = 1 | G | X λ irred. d λ tr ( b µ ( λ ) b µ ( λ ) T ) . This establishes the connection with the total variation distance if we take µ on the left hand side to be the difference of two measures.On the right hand side, we have the trace of Fourier transforms, whichcan be evaluated by writing it in terms of the characters of the irreduciblerepresentations. We denote the character of the representation λ by χ λ , which is defined as χ λ ( g ) = tr ( λ ( g )) . An important fact about characters isthat they are class measures, so they are constant on conjugacy classes. Wehave the following lemma, which can be found in Chapter 16 of [2].
ALPEREN Y. ¨OZDEMIR
Lemma 2.1. ([2])
Let a class measure µ be defined on G and λ be a repre-sentation of G. Then, b µ ( λ ) = 1 d λ X g ∈ G µ ( g ) χ λ ( g ) I d λ . We also note a fact about the Fourier transform of the convolution ofmeasures, as a Markov chain at step t can be viewed as t − fold convolutionmeasure of the transition probabilities. The convolution of two measures µ and ν is defined to be µ ∗ ν ( g ) = X h ∈ G µ ( gh − ) ν ( h )The Fourier transform of a convolution satisfies(4) [ µ ∗ ν ( g ) = b µ ( g ) b ν ( g ) . Lower bound.
It follows from the definition of the total variationdistance (1), after a random ( n − cn transpositions, we have(5) k µ cn +1 − U cn +1 k T V ≥ | µ cn +1 ( A ) − U cn +1 ( A ) | where A is the set of fixed point free permutations. This particular choiceof the subset of S n establishes a connection with the symmetric group rep-resentations.We first introduce the defining representation of S n , which is the n -dimensional representation ρ where ρ ( σ )( i, j ) = (cid:26) σ ( j ) = i σ ∈ S n . Denote the character of the representation, or the traces of the matri-ces given above, by χ ρ . So that χ ρ ( σ ) counts the number of fixed points of σ ∈ S n . Therefore σ ∈ A if and only if χ ρ ( σ ) = 0 . The number of fixed points of a uniformly random permutation has lim-iting Poisson distribution with parameter 1; elementary proofs of this factcan be found in [19]. The limiting distribution in our case, which alternatesbetween the uniform distribution over permutations in A n and in S n \ A n , isalso asymptotically P (1). A proof is given in [10] by showing that the r th moment of χ ρ under U t agrees with the r th moment of P (1) for r ≤ n, i,e., E U t ( χ rρ ) = r X i =0 (cid:26) ri (cid:27) = B r for r ≤ n, where (cid:8) ri (cid:9) is a notation for the Stirling numbers of second kind.The sum above gives the r th Bell number.
HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP Next we consider the moments of χ ρ under the law of the Markov chainat step t. The defining representation is reducible and decomposed as(6) ρ = S ( n ) ⊕ S ( n − , where S λ is the Specht module associated with partition λ of n , noting thatthe partitions of n are in one to one correspondance with the irreduciblerepresentations of S n . See Chapter 4 of [12] for details.In order to find the higher moments, we consider the decomposition ofthe tensor product ρ ⊗ r .(7) ρ ⊗ r = ⊕ λ ⊢ n a λ,r S λ where λ ⊢ n means that λ is a partition of n . Then we use the facts below, χ ρ ⊕ ρ = χ ρ + χ ρ ,χ ρ ⊗ ρ = χ ρ · χ ρ , and invoke some Fourier analytic results, which can be found in [2] Chapter16 with detailed proofs and in [7] Section 2C in most relevance to our case,to calculate the moments needed.(8) E µ (( χ ρ ) r ) = X λ ⊢ n a λ,r tr (ˆ µ ( λ )) . Now define ¯ λ = ( λ , λ , ..., λ m ), the partition obtained by removing thefirst row of λ . The generating function of coefficients a λ,r for a fixed partition λ , X r ≥| ¯ λ | a λ,r x r r ! = d λ | ¯ λ | ! e e x − ( e x − | ¯ λ | , is found by Goupil and Chauve in [11]. The coefficients are identified byDing in [10] to be(9) a λ,r = d ¯ λ r X i = | ¯ λ | (cid:18) i | ˆ λ | (cid:19)(cid:26) ri (cid:27) . for 1 ≤ r ≤ n − λ . Therefore, explicit expressions for higher moments of χ ρ can be obtained by finding the traces of the Fourier transforms of µ. Lemma2.1 and the fact about the convolution of measures (4) are applied in thatrespect. Particularly in our case,tr( ˆ µ t ( λ )) = d λ χ λ ( n − k, k ) d λ ! χ λ (2 , n − ) d λ ! t where χ λ ( n − k, k ) is the character of the representation λ evaluated at an( n − k ) − cycle.Very often it might be difficult to calculate the the higher moments for anarbitrary permutation. The following lower bound lemma requires the first ALPEREN Y. ¨OZDEMIR moment and a higher one to provide a lower bound on the total variationdistance. We follow the proof of a result in Chapter 7 of [14], which gives alower bound using the first two moments.
Lemma 2.2.
Let µ and ν be two probability distributions on Ω , and X bea random variable from Ω to a finite subset A of R . Then for p, q ∈ (1 , ∞ ) satisfying p + q = 1 , k µ − ν k T V ≥
12 [ E µ ( X ) − E ν ( X )] q ( E µ ( X p ) + E ν ( X p )) q − . Proof:
We first define the average measure on A , σ ( x ) = P µ ( X = x ) + P ν ( X = x )2 , and the functions α ( x ) = P µ ( X = x ) σ ( x ) , β ( x ) = P ν ( X = x ) σ ( x ) . Then by H¨older’s inequality, E µ ( X ) − E ν ( X ) = X x ∈ A x ( α ( x ) − β ( x )) σ ( x ) ≤ X x ∈ A x p σ ( x ) ! /p X x ∈ A | α ( x ) − β ( x ) | q σ ( x ) ! /q (10)where p + q = 1 for p, q ∈ (1 , ∞ ) . Observe that X x ∈ A x p σ ( x ) = E µ ( X p ) + E ν ( X p )2 , X x ∈ A | α ( x ) − β ( x ) | σ ( x ) = 2 k µ − ν k T V and | α ( x ) − β ( x ) | = 2 | P µ ( X = x ) − P ν ( X = x ) | P µ ( X = x ) + P ν ( X = x ) ≤ . Combining the observations above in (10), we arrive at the result. (cid:3)
Upper bound.
An upper bound lemma is provided in Chapter 3B of[7]. Yet we need a slight modification of it due to the alternation of thestationary distribution U t between A n and S n \ A n in our case.We first introduce the notation for the two one dimensional irreduciblerepresentations of S n . The first one is λ triv is the trivial representation thatmaps every element to 1. The other one is λ sign , which is 1 for even per-mutations and − HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP Lemma 2.3.
Let µ t be the law of the Markov chain at step t and U t be thestationary distribution described in (1). Then, k µ t − U t k T V ≤ X λ irred.λ = λ triv ,λ sign d λ tr ( b µ t ( λ ) b µ t ( λ ) T ) . Proof:
We first argue for the following facts on the Fourier transform of U t . c U t ( λ ) = b µ t ( λ ) if λ = λ triv or λ sign , c U t ( λ ) is the zero matrix otherwise.(11)Clearly, c U t ( λ triv ) = 1 . c U t ( λ sign ) is either 1 or − t. Suppose that the chain is on A n . Then b µ t ( λ sign ) = X σ ∈ A n µ t ( σ ) χ λ sign ( σ ) = X σ ∈ A n µ t ( σ ) = 1 , which is equal to c U t ( λ sign ). If the chain on S n \ A n , they are both − . If λ is different from the trivial and the sign representation, c U t ( λ ) is the zerotransformation as a consequence of Schur’s lemma. See Chapter 2B of [9]for Schur’s lemma applications in this context.The first step below follows from Cauchy-Schwarz inequality noting that µ t and U t have support either A n or its coset depending on t . The secondfollows from Plancherel’s formula (3) and (11).4 k µ t − U t k T V = X σ ∈ S n | µ t ( σ ) − U t ( σ ) | ! ≤ n !2 X σ ∈ S n | µ t ( σ ) − U t ( σ ) | = 12 X λ irred.λ = λ triv ,λ sign d λ tr ( b µ t ( λ ) b µ t ( λ ) T ) (cid:3) We evaluate the sum above even further by Lemma 2.1 and (4) to arriveat(12) 4 k µ t − U t k T V ≤ X λ irred.λ = λ triv ,λ sign d λ χ λ ( n − k, k ) d λ ! χ λ (2 , n − ) d λ ! t . 3. Proof of Theorem 1.1
The moment sequence of χ ρ . We find the moments of χ ρ by theformula (8). We start with evaluatingtr( b µ ( λ )) = d λ χ λ ( n − , d λ ! χ λ (2 , n − ) d λ ! t ALPEREN Y. ¨OZDEMIR for λ ⊢ n. Observe that for most of the partitions of n , the character evalu-ated at ( n − − cycle gives 0 by the Murnaghan-Nakayama rule (See Chapter4 of [18] for the Murnaghan-Nakayama rule). Diaconis and Greene [8] cal-culated the characters for the remaining partitions, χ λ ( n − , = λ = ( n )( − | ¯¯ λ | +1 λ = ( n − − i, , i )0 otherwisewhere ¯¯ λ = ( λ , λ , ..., λ m ). Furthermore Ding [7] has the following resultsfor the character of λ = ( n − − i, , i ) evaluated at transpositions, χ λ (2 , n − ) d λ = 1 − in for i ≤ (cid:4) n − (cid:5) . Therefore we have asymptotically,(13) tr( \ µ cn +1 ( λ )) ∼ ( − | ¯¯ λ | +1 e − c (2+ | ¯¯ λ | ) for λ = ( n − − i, , i ) and i ≤ (cid:4) n − (cid:5) . To compute the moments, we need the following auxillary fact.
Lemma 3.1.
For x ∈ R , n X k =0 k + 2 (cid:18) nk (cid:19) ( − x ) k +2 = (1 − x ) n +2 − n + 2 − (1 − x ) n +1 − n + 1 . Proof:
Take the derivative of the left hand side with respect to x. Then usebinomial theorem to have x (1 − x ) n . Next integrate x (1 − x ) n , and solve forthe integral constant considering that the expression on the left hand sideis 0 for x = 0 . (cid:3) All needed to calculate the moments of χ ρ with respect to µ cn +1 arederived. First observe that E µ cn +1 (( χ ρ )) = 1 , since ρ = S ( n ) ⊕ S ( n − , and χ ( n − , n − , = 0 . For the higher moments, combine (9) and (13) in the formula 8 for 2 ≤ r ≤ (cid:4) n − (cid:5) to arrive at HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP E µ cn +1 (( χ ρ ) r ) = a ( n ) ,r + X λ ⊢ n a λ,r tr ( \ µ cn +1 ( S λ )) ∼ r X i =1 (cid:26) ri (cid:27) + n − X k = | ¯¯ λ | =0 r X i = | ¯ λ | = k +2 ( k + 1) (cid:18) ik + 2 (cid:19)(cid:26) ri (cid:27) ( − k +1 e − c ( k +2) = r X i =1 (cid:26) ri (cid:27) + r X i =2 i − X k =0 ( k + 1) (cid:18) ik + 2 (cid:19)(cid:26) ri (cid:27) ( − k +1 e − c ( k +2) =1 + r X i =2 (cid:26) ri (cid:27)(cid:16) − r X i = k +2 ( k + 1) (cid:18) ik + 2 (cid:19) ( − e − c ) k +2 (cid:17) =1 + r X i =2 (cid:26) ri (cid:27)(cid:16) − i ( i − r X i = k +2 k + 2 (cid:18) i − k (cid:19) ( − e − c ) k +2 (cid:17) . Then apply Lemma 3.1 for x = e − c to have E µ cn +1 (( χ ρ ) r ) ∼ r X i =2 (cid:26) ri (cid:27)(cid:16) − i ( i − (cid:16) (1 − e − c ) i − i − (1 − e − c ) i − − i − (cid:17)(cid:17) =1 + r X i =2 (cid:26) ri (cid:27)(cid:16) (1 − e − c ) i + ie − c (1 − e − c ) i − (cid:17) = r X i =1 (cid:26) ri (cid:27)(cid:16) i e − c − e − c (cid:17) (1 − e − c ) i . (14)One observation, which is made earlier in [7], is that the moments ob-tained above is pretty close to the moments of the Poisson distribution withparameter (1 − e − c ) . In fact, Kuba and Panholzer [13] shows that if themiddle term in the sum above coincided with the i th moment of a distri-bution, we would have a mixed Poisson random variable where the Poissonparamater has moments proportional to the middle term. But one can eas-ily show that the middle term does not qualify to be a moment sequence.However, Proposition 2 in [13] on the distribution related to the momentsequence of a mixed Poisson distribution can easily be modified to our case,which is in the next section.3.2. The asymptotic distribution of χ ρ . In this section, we determinethe asymptotic distribution of χ ρ by its moment sequence. We start with awell-known theorem that gives a sufficient condition for the uniqueness of adistribution given its moment sequence. Theorem 3.1. ([5])
Let X be a real random variable having finite moments µ n . If the moment generating function E ( e zX ) of X has positive radius ofconvergence, then the distribution of X is the only distribution with themoment sequence µ n . A combinatorial fact that is used for evaluating the sums below is asfollows.
Lemma 3.2. ([1]) If n > m , n X k =0 ( − n − k (cid:18) nk (cid:19) k m = 0 . Next we verify the hypothesis of Theorem 3.1 for χ ρ . Proposition 3.1.
The moment generating function of χ ρ with distribution µ cn +1 has positive radius of convergence.Proof: Let ζ := 1 − e − c and η := e − c − e − c in (14). Then, E ( e zχ ρ ) = ∞ X i =0 E µ cn +1 (( χ ρ ) i ) z i i != ∞ X i =0 i X j =0 (cid:26) ij (cid:27) ζ j (1 + jη ) z i i ! ≤ ∞ X i =0 i X j =0 (cid:26) ij (cid:27) π j (1 + η ) j z i i !First applying Lemma 3.2, then using the two-variable generating functionidentity involving the Stirling numbers of the second kind, which can befound in Chapter 3 of [20], we obtain E ( e zχ ρ ) ≤ ∞ X i =0 ∞ X j =0 (cid:26) ij (cid:27) ζ j (1 + η ) j z i i ! = e ζ (1+ η )( e z − . (cid:3) Therefore, it is proven by Theorem 3.1 and Proposition 3.1 that the mo-ment sequence found above uniquely determines the distribution.The next is a variation of the result discussed above, Proposition 2 in [13].
Proposition 3.2.
Let X denote a random variable with probability massfunction, P ( X = j ) = ∞ X i = j ( − i − j (cid:18) ij (cid:19) α i β i i ! . HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP Then for all r ∈ N , the moments of X is given by E ( X r ) = r X i =1 (cid:26) ri (cid:27) α i β i . Proof:
The r th moment of X can be expressed as E ( X r ) = ∞ X j =1 ∞ X i = j ( − i − j (cid:18) ij (cid:19) α i β i i ! j r = ∞ X i =1 i X j =1 ( − i − j (cid:18) ij (cid:19) α i β i i ! j r . Then by Lemma 3.2, E ( X r ) = ∞ X i =1 (cid:26) ri (cid:27) α i β i i != r X i =1 (cid:26) ri (cid:27) α i β i i ! . (cid:3) Finally, we identify the distribution of χ ρ as n goes to infinity for thesymmetric group S n . If we take α i to be (cid:16) i e − c − e − c (cid:17) and β to be 1 − e − c in Proposition 3.2, we have(15) P ( χ ρ = j ) ∼ ∞ X i = j ( − i − j (cid:18) ij (cid:19)(cid:16) i e − c − e − c (cid:17) (1 − e − c ) i i ! , following from Theorem 3.1 and Proposition 3.1.3.3. Lower bound for the (n-1)-cycle case.
We first calculate the prob-ability that the trace of the defining representation after one ( n − − cycle and cn transpositions is 0. Taking j = 0 in (15), P ( χ ρ = 0) = ∞ X i =0 ( − i (cid:18) i (cid:19)(cid:16) i e − c − e − c (cid:17) (1 − e − c ) i i ! ∼ ∞ X i =0 ( − i (1 − e − c ) i i ! + ∞ X i =0 ( − i i e − c − e − c (1 − e − c ) i i != e − (1 − e − c ) + ∞ X i =0 ( − i i e − c − e − c (1 − e − c ) i i != e − (1 − e − c ) − e − c ∞ X i =0 ( − i (1 − e − c ) i i != e − (1 − e − c ) − e − c e − (1 − e − c ) = e e − c − (1 − e − c ) . Then take A to be the set of fixed point free permutations, since χ ρ countsthe number of fixed points,(16) µ cn +1 ( A ) = P ( χ ρ = 0) ∼ e e − c (1 − e − c ) e . Finally plug it in the lower bound inequality (5). k µ cn +1 − U cn +1 k T V ≥| µ cn +1 ( A ) − U cn +1 ( A ) |∼ (cid:12)(cid:12) e e − c (1 − e − c ) e − e (cid:12)(cid:12) = 1 − e e − c + e e − c − c e . The proof of Theorem 1.1 is completed with the result for the upper boundin [7]. (cid:3)
Remark 3.1.
Unlike the n -cycle case, which is studied in [10] , we have µ cn +1 ( A ) < U cn +1 ( A ) for the ( n − − cycle case. The comparison is madeby the series expansion of the term above, which is ∞ X i =0 i + 2 ( e − c ) i i ! > . Proof of Theorem 1.2
Lower bound for (n-k)-cycle case for k ≥ . The moment sequenceof χ ρ for any k ≥ χ ρ unlike k = 1 case above. However, we can derive a lowerbound through the first two moments for k ≥
3. The third moment is usedonly for k = 2 case. The reason is that the the third moment calculationsare cumbersome for k ≥ HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP of the lower bound. But for k = 2, both the calculations are relatively easyand the constant in the lower bound is significantly larger.We have above the decomposition of ρ (6), and by formula (7) we canfind the decomposition of ρ ⊗ ρ to be(17) ρ ⊗ ρ = 2 S ( n ) ⊕ S ( n − , ⊕ S ( n − , ⊕ S ( n − , ) . Therefore, we have the following expressions for the first two moments. E µ t ( χ ρ ) = tr( ˆ µ t ( n )) + tr( ˆ µ t ( n − , E µ t ( χ ρ ) = 2 tr( ˆ µ t ( n )) + 3 tr( ˆ µ t ( n − , µ t ( n − , µ t ( n − , , U t ∼ P (1) , we also have E U t ( χ ρ ) ∼ E U t ( χ ρ ) ∼ χ ( n )( n − k, k ) = 1 for all kχ ( n − , n − k, k ) = k − k ≥ χ ( n − , n − k, k ) = − k = 20 k = 31 + ( k )( k − k ≥ χ ( n − , )( n − k, k ) = k = 21 k = 32 + ( k +1)( k − k ≥ The characters on transpositions are evaluated in Chapter 3 of [7], we presentthem below and find the asymptotics for t = ln k n + cn.χ ( n )(2 , n − ) d ( n ) = 1 χ ( n − , , n − ) d ( n − , t = (cid:16) n − n − (cid:17) t ∼ k e − c χ ( n − , , n − ) d ( n − , t = (cid:16) n − n − (cid:17) t ∼ k e − c χ ( n − , )(2 , n − ) d ( n − , ) t = (cid:16) n − n (cid:17) t ∼ k e − c (21)So we can write down the moments by evaluating the expressions in (18). E µ t ( χ ρ ) ∼ k − k e − c for all k ≥ E µ t ( χ ρ ) ∼ e − c − e − c k = 22 + 2 e − c + e − c k = 32 + 3 k − k e − c + k − k +1 k e − c k ≥ k ≥ p = q = 2in Lemma 2.2, k µ t − U t k T V ≥ (cid:0) k − k e − c (cid:1) k − k e − c + k − k +1 k e − c ≥ (cid:0) k − k (cid:1) k − k + k − k +1 k e − c ≥ e − c , if k ≥ c = 0 in the denominator for the second inequality, since c can be arbitrarily close to 0. For the last inequality k is taken 4, whichminimizes the expression.By the same method, we find a lower bound for k = 3 . k µ t − U t k T V ≥ (cid:0) e − c (cid:1) e − c + e − c ≥ e − c , if k = 3(24)For k = 2, we consider the third moment of χ p since the third momentyields a constant twice larger the one obtained by the second moment. We HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP have the decomposition of ρ ⊗ by (7), which is ρ ⊗ = 5 S ( n ) ⊕ S ( n − , ⊕ S ( n − , ⊕ S ( n − , ) ⊕ S ( n − , ⊕ S ( n − , , ⊕ S ( n − , ) . By Murnaghan-Nakayama rule, we evaluate the characters of the represen-tations in the decomposition above, χ λ ( n − , ) = − λ = ( n − , λ = ( n − , , λ = ( n − , )Next we evaluate the non-vanishing character at transpositions and find theasymptotics for t = ln k n + cn. χ ( n − , , n − ) d ( n − , t = (cid:16) ( n − n − n ( n − (cid:17) t ∼ k e − c . Therefore together with (20) and (21), we have E µ t ( χ ρ ) = 5 + 5 e − c − e − c − e − c for k = 2 . Finally apply Lemma 2.2 taking p = 3 and considering that E U t ( χ ρ ) ∼ k µ t − U t k T V ≥ (cid:0) e − c (cid:1) / (cid:0)
10 + 5 e − c − e − c − e − c (cid:1) / ≥ / √ e − c ≥ e − c , if k = 2 . (25)We compare the bounds in (23), (25) and (25) to see that e − c is thesmallest, which proves the left hand side of the expression in Theorem 1.2. Remark 4.1.
The higher moments can provide better lower bounds. How-ever, the calculations are quite lengthy and it is not essential for our purpose.Therefore the third moment calculations for k ≥ are skipped. Auxillary facts for the upper bound.
The rest of the paper is onthe upper bound in Theorem 1.2. We first define a central object for thefollowing sections. Then we state two important lemmas which play key rolein the evaluation of (12).
Definition 4.1.
A skew diagram of a Young diagram is a subset of cellssuch that when removed the diagram obtained is a Young diagram. A rimhook (also known as ribbon) of a Young diagram is an edgewise connectedskew diagram with no × cells. See Figure 1 for non-examples and an example of a rim hook.(a) (b) a rim hook
Figure 1.
The shaded region on (a) fails to be a skew dia-gram. While the shaded region on (b) is a skew diagram, itis not edgewise connected.
Lemma 4.1. ([15])
For any Young diagram of shape λ ⊢ n there exists atmost one way to remove a rim hook of length m > n . A purely combinatorial proof of Lemma 4.1 is found in Section 6 of [15].
Lemma 4.2. ([7])
Let | λ | = n and λ be the size of the first row of λ. Then, X λ =( λ ,... ) d λ ≤ (cid:18) nλ (cid:19) ( n − λ )! . The result is stated in Section 3D of [7]. It follows from two results statedearlier in [7]. One is a corollary to Fact 1 in Section 3D, which gives a boundon d λ . The second one is Corollary 1 in Chapter 2C, which is an importantfact in representation theory that the sum of the squares of the dimensionsof irreducible representations is equal to the order of the group.4.3. Estimates for the normalized character of transpositions.
Inthis section, we present estimates for normalized characters, which lead toa monotonicity result that is used to estimate the upper bound on the totalvariation distance in 4.4.
Lemma 4.3. ([7])
Let λ = ( λ , λ , ..., λ m ) be a partition of n and (2 , n − ) stands for transpositions. Then we have the following formula. r ( λ ) := χ λ (2 , n − ) d λ = 1 n ( n − m X i =1 (cid:16) λ i − (2 i − λ i (cid:17) = 1 (cid:0) n (cid:1) m X i =1 (cid:18) λ i (cid:19) − (cid:18) λ Ti (cid:19) . (26) Lemma 4.4.
Let λ = ( λ , λ , ..., λ m ) be a partition of n . Then, r ( λ ) ≤ λ − λ ( λ − λ +2) n n − . HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP Proof:
By Lemma 4.3, n ( n − r ( λ ) = X λ i ( λ i − (2 i − λ ( λ −
1) + λ ( λ −
3) + · · ·
Then we can bound the expression above as n ( n − r ( λ ) ≤ λ ( λ −
1) + ( λ − X i =2 λ i = λ ( λ − λ + 2) + λ ( λ −
3) + ( λ − X i =2 λ i =( λ − n + λ ( λ − λ + 2) . (cid:3) We define the following partial order on the set of partitions of n , called dominance order , which is used to compare the normalized characters oftranspositions in the proof of the upper bound. Definition 4.2. ( [16] ) If λ = ( λ , λ , ... ) and ξ = ( ξ , ξ , ... ) are partitionsof n, we say that λ dominates ξ , and denote by λ (cid:23) ξ, if j X i =1 λ i ≥ j X i =1 ξ i for all j.Besides, if λ (cid:23) ξ and λ = ξ, then we write λ ≻ ξ. The following result is in Chapter 3D of [7].
Lemma 4.5. ([7]) If λ (cid:23) ξ, then r ( λ ) ≥ r ( ξ ) . An interpretation of the partial order defined above is that λ is obtainedfrom ξ by moving boxes up to the right [7]. The following lemma is crucialfor evaluating the sums in the next section. Lemma 4.6.
Suppose λ is obtained from λ ′ by moving a box on the rim upto the right. i.e., λ ′ = ( λ , λ , ..., λ k − , λ k − , λ k +1 , ..., λ l − , λ l + 1 , λ l +1 , , ... ) for some k < l and λ l < λ k . Then r ( λ ) − r ( λ ′ ) is n ( n − times the numberof boxes on the rim between the two positions of the displaced box.Proof: We calculate the difference in the normalized characters by Lemma4.3. r ( λ ) − r ( λ ′ ) = 1 n ( n − X (cid:16) λ i − (2 i − λ i (cid:17) − (cid:16) λ ′ i − (2 i − λ ′ i (cid:17)! = 1 n ( n − (cid:16) λ k ( λ k − (2 k − − ( λ k − λ k − k ) (cid:17) + 1 n ( n − (cid:16) λ l ( λ l − (2 l − − ( λ l + 1)( λ l − (2 l − (cid:17) = 2( λ k − λ l + ( l − k ) − n ( n − . (cid:3) Figure 2 below illustrates an example for Lemma 4.6. • •• λ = (6 , , , , λ ′ = (6 , , , , Figure 2.
A pair of Young diagrams that satisfy the hypoth-esis of Lemma 4.6. We have r λ − r λ ′ = n ( n − × × = . Remark 4.2.
A more general statement for Lemma 4.6, along with variousestimates for normalized characters, is found in [4] . But Lemma 4.6 issufficient for our purpose, to limit the growth of successive terms in theupper bound in the next section.
Upper bound for (n-k)-cycle case for k ≥ . This final section isdevoted to simplifying the sum in (12) and bounding it from above. We firstexpress it with the normalized character notation.(27) 4 k µ t +1 − U t +1 k T V ≤ X λ ⊢ nλ = λ triv ,λ sign ( χ λ ( n − k, k ) ) r ( λ ) t . Consider the first term, ( χ λ ( n − k, k ) ) , in the sum. Since k is fixed, eventually( n − k ) > n . There is thus exactly one way to remove a rim hook of length( n − k ) by Lemma 4.1. An easy corollary to Murnaghan-Nakayama rulegives ( χ λ ( n − k, k ) ) = ± d ξ where ξ is the unique diagram obtained from λ by removing the rim hookof length ( n − k ) . Then we write the sum (27) over the partitions obtainedafter the removal of a rim hook.(28) 4 k µ t +1 − U t +1 k T V ≤ X | ξ | = kξ =( k ) , (1 k ) n − k X s =1 d ξ r ( λ sξ ) t + n − k X s =2 d k ) r ( λ s ( k ) ) t + n − k − X s =1 d k ) r ( λ s (1 k ) ) t HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP λ sξ in the sum denotes a partition of n that yields ξ after removing the rimhook, and s is an index for an ordering on the vertical position of the down-most cells of the rim hook among all such partitions. See Figure 3 below. If n is large enough, there are exactly ( n − k ) possible ways to recover a Youngdiagram of size n by attaching a rim hook of size ( n − k ) to a diagram ofsize k . Therefore, s runs through 1 to n − k, where λ ξ = ( n − k, k ) . Thesecond sum lacks s = 1 , because it corresponds to the trivial representation.Similarly, the sign representation is excluded in the third sum. Note that if k = 2 , then the first tem in the paranthesis above vanishes.1 2 3 . . . n − n − s = 1 and λ ξ = ( n − , , . . . . n − n − s = 2 and λ ξ = ( n − , , . ... n − n − n − n − n − ...21 n − n − ...21 s = n − λ n − k − ξ = (2 , n − ) . s = n − λ n − kξ = (2 , n − ) . Figure 3.
Young diagrams of λ ξ , ..., λ n − kξ for k = 4 and ξ = (2 , , . We regroup the terms in (28) finer. It suffices to consider the partitionswith ξ ≥ ξ T where ξ T is the transpose diagram of ξ and ξ is the size of the first row of ξ. The justification comes from a symmetric and an anti-symmetric relation below, d λ = d λ T r ( λ ) = − r ( λ T ) by (26) . (29)The second and the third sum in (28) are the same as well by symmetry andthe fact that d ( k ) = d (1 k ) = 1. Therefore we have,4 k µ t +1 − U t +1 k T V ≤
12 2 k − X i = ⌈√ k ⌉ X | ξ | = kξ = i n − k X s =1 d ξ r ( λ sξ ) t + n − k − X s =1 r ( λ sξ ) t = k − X i = ⌈√ k ⌉ X | ξ | = kξ = i d ξ n − k X s =1 r ( λ sξ ) t + n − k − X s =1 r ( λ sξ ) t . (30)Next we evaluate the sums n − k P s =1 r ( λ sξ ) t and n − k − P s =1 r ( λ s ( k ) ) t . We identifythe largest term in the sum and the difference between the successive terms,and then we bound it from above. Take i = ξ . The first observation is that λ ξ ≻ λ ξ ≻ · · · λ n − k − ξ ≻ λ n − kξ . So by Lemma 4.5, the maximum term of the sum is achieved for s = 1. Thebound for that term is derived from Lemma 4.4 below. r ( λ ξ ) ≤ i − ( n − k + i )( n − k +2) n n − n − k + 2 i − o (1) n − − k − i ) + o (1) n − r ( λ ξ ), i.e., − r ( λ n − kξ ) = r (( λ n − kξ ) T ) ≤ i − ( n − k + i )( n − k +2) n n − − k − i ) + o (1) n − . (32)by (29)Finally, it follows from Lemma 4.6 that for the successive diagrams we have(33) r ( λ iξ ) − r ( λ i +1 ξ ) ≥ n − k − n ( n −
1) = 2( n − k ) − n ( n −
1) = 2 − o (1) n − . Lemma 4.6 is applicable above. Because λ i +1 can be thought to be obtainedfrom λ i by moving the box on the rightmost and uppermost end of the rim HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP hook of length ( n − k ) to the leftmost and downmost end, if it results in an-other Young diagram. If not, it is moved even further left or further below.Therefore the number of boxes on the rim between two positions is greaterthan or equal to n − k − p to be the largestindex for which r ( λ p ) is positive in order to split the inner sum for negativeand positive normalized characters; the justification is by the symmetry (32)observed above. Then by (31) and (33), we have(34) p X s =1 r ( λ sξ ) t ≤ p − X s =0 (cid:16) − k − i ) + 2 s + o (1) n − (cid:17) t Taking t = cn + n ln k in (34), p X s =1 r ( λ sξ ) t ≤ p − X s =0 (cid:16) − k − i ) + 2 s + o (1) n − (cid:17) cn + n ln k ∼ p − X s =0 e − k − i + s ) c k − k − i + s ) . (35)Similarly, n − k X s = p +1 r ( λ sξ ) t ≤ n − k − p − X s =0 (cid:16) − k − i ) + 2 s + o (1) n − (cid:17) t ∼ p − X s =0 e − k − i + s ) c k − k − i + s ) . (36)We add (35) and (36) below to arrive at an upper bound for the sum. n − k X s =1 r ( λ sξ ) t ≤ ∞ X s =0 e − k − i + s ) c k − k − i + s ) =2 e − k − i ) c k − k − i ) ∞ X s =0 ( e − c k − ) s = 21 − e − c k − e − k − i ) c k − k − i ) . The argument above also works for n − k P s =2 r ( λ s ( k ) ) t , the second term in (26).Consider λ k ) ≻ λ k ) ≻ · · · λ n − k − k ) ≻ λ n − k ( k ) , where λ k ) = ( n − k − , k + 1) and λ n − k ( k ) = ( k, n − k ). Apply Lemma 4.4 tohave r ( λ k ) ) ≤ − k + 2 + o (1) n − , − r ( λ n − k ) ) ≤ − k − o (1) n − . So the leading term in absolute value is bounded by 1 − k − i )+ o (1) n − . Werepeat (35) and (36), and conclude n − k X s =2 r ( λ s ( k ) ) t ≤ − e − c k − e − k − c k − k − . Going back to the upper bound inequality (30), we replace the sum ofnormalized characters in the expression by the bounds found for them above.4 k µ t +1 − U t +1 k T V ≤ − e − c k − k − X i = ⌈√ k ⌉ e − k − i ) c k − k − i ) X | ξ | = kξ = i d ξ + e − k − c k − k − ≤ − e − c k − k − X i = ⌈√ k ⌉ (cid:18) ki (cid:19) ( k − i )! e − k − i ) c k − k − i ) + e − k − c k − k − The second inequality follows from Lemma 4.2. We change the index from i to j = k − i for simplicity, and simplify the expression even further. For k ≥ c > k µ t +1 − U t +1 k T V ≤ − e − c k − ⌊ k −√ k ⌋ X j =1 (cid:18) kj (cid:19) j ! e − jc k − j + e − k − c k − k − = 21 − e − c k − ⌊ k −√ k ⌋ X j =1 j ! k · · · ( k − j + 1) k j e − j − c + e − k − c k − k − e − c ≤ − k − ⌊ k −√ k ⌋ X j =1 j ! k · · · ( k − j + 1) k j + k − k − e − c ≤ ∞ X j =1 j ! e − c =2( e − e − c . HE RANDOM ( n-k )-CYCLE TO TRANSPOSITIONS WALK ON THE SYMMETRIC GROUP This, combined with the lower bound result in 4.1, completes the proof ofTheorem 1.2. (cid:3)
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