Permutation statistics of products of random permutations

0.15 MB·English

by  Axel Hultman

PERMUTATION STATISTICS OF PRODUCTS OF RANDOM
PERMUTATIONS
AXELHULTMAN
Abstract. Givenapermutationstatistics:Sn→R,definethemeanstatis-
ticsasthestatisticwhichcomputesthemeanofsoverconjugacyclasses. We
3 describeawaytocalculatetheexpectedvalueofsonaproductoftindepen-
1 dentlychosenelementsfromtheuniformdistributiononaunionofconjugacy
0 classesΓ⊆Sn. Inordertoapplytheformula,oneneeds toexpress theclass
2 function s as a linear combination of irreducible Sn-characters. We provide
n suchexpressions forseveral commonlystudiedpermutation statistics, includ-
a ing the excedance number, inversion number, descent number, major index
J and k-cycle number. In particular, this leads to formulae for the expected
valuesofsaidstatistics.
3

]
O
C 1. Introduction
.
h Consider the symmetric group S of permutations of [n] = {1,...,n}. For
n
at Γ⊆Sn one may study the behaviour of various permutation statistics s:Sn →R
m on products γ ···γ ∈S of random γ ∈Γ.
1 t n i
[ Definition1.1. ChooseasubsetΓ⊆Sn,afunctions:Sn →Randanonnegative
1 integer t. We denote by E (s,t) the expected value of s on a product of t elements
Γ
v independently chosen from the uniform distribution on Γ.
0
3 A product of t random elements of Γ corresponds to a t-step random walk on
4 the Cayley graph of S induced by Γ. Random walks on Cayley graphs form a
n
0
classical and well studied subject in probability theory; a good general reference
.
1 is [12]. In the present paper, we specifically address the problem of computing
0 E (s,t). Recent work in this vein includes the following. The case of Γ being the
3 Γ
set of adjacenttranspositions and s counting inversionswas studied by Erikssonet
1
: al. [4], Eriksen[5] and Bousquet-M´elou[2]. Turning, instead, to Γ comprisedof all
v transpositions, s being the absolute length function (essentially counting disjoint
i
X cycles) was studied in [6], whereas J¨onsson [9] considered the fixed point counting
r function s and Sjo¨strand [13] found the solution when s counts inversions.
a We shall describe a method to attack the general problem. Although it could
potentiallybeofuseformoregeneralΓ(asindicatedbythehypothesesofTheorem
3.1 below), we shall apply it to situations where Γ is a union of conjugacy classes.
The technique makes use of representation theory of S . Similar ideas have been
n
frequentinthestudyofrandomwalksonCayleygraphssincetheseminalpaperby
DiaconisandShahshahani[3]. Inparticular,ourmethodisheavilyinspiredbythat
described in [6]. We are concerned with more general Γ, but the principal novelty
here is to dispose of the apparent requisite of [6] that s be a class function. As
we shall see, removing this restriction significantly improves the versatility of the
method.
Here is a brief sketchofthe content ofthe paper. In the next section, we review
some facts about symmetric group characters. In Section 3, we then describe how
theyconnectwithourexpectedstatisticsproblem. Themainresultsofthatsection,
Theorem3.1 andTheorem3.2, provide a recipe for computing E (s,t) whenever Γ
Γ
1
2 AXELHULTMAN
is a union of conjugacy classes. In order to express E (s,t) explicitly for a given
Γ
statistic s,the remainingtask isto decomposethe mean statistics asalinearcom-
binationofirreducibleS -characters. This turns outto be afairly straightforward
n
taskformanystandardpermutationstatistics. We provideexplicitdecompositions
for the mean statistics corresponding to the k-cycle number, excedance number,
weak excedance number, inversion number, major index and descent number sta-
tistics in Sections 4, 5 and 6. In particular, E (s,t) is determined for s being any
Γ
of these statistics (and conjugation invariant Γ). We conclude with a few explicit
examples in Section 7.
2. Symmetric group characters
Inthissection,wereviewelementsoftherepresentationtheoryofS . Fromthis
n
vast and classical subject only a few bits and pieces that we need in the sequel are
extractedinordertoagreeonnotation. Wereferto[11]forathoroughbackground
and much more information.
LetP denotethesetofintegerpartitionsλ⊢n. Theirreduciblerepresentations
n
of S are in bijection with P in a standard way. We use the notation ρλ for the
n n
representation indexed by λ ⊢ n and denote the corresponding character by χλ.
TheseirreduciblecharactersformabasisfortheC-vectorspaceCl ={f :P →C}
n n
ofclassfunctions. Moreover,thisbasisisorthonormalwithrespecttothe standard
Hermitian inner product on Cl
n
1
hf,gi= |C |f(λ)g(λ)∗,
λ
n!
λ⊢n
X
where C denotes the conjugacy class of permutations π with type(π)=λ as cycle
λ
type.
Abusing notation, we at times consider class functions as defined on S rather
n
than on P . In other words, for π ∈ S and f ∈ Cl , f(π) should be interpreted
n n n
as f(type(π)). We trust the context to prevent confusion.
It is convenient to encode partitions as weakly decreasing sequences of positive
integers, sometimes employing exponent notation to signal repeated parts. For
example, (7,34,1) denotes the partition of 20 which consists of one part of size
7, four parts of size 3 and one part of size 1. In this notation, a hook shape is a
partition of the form (a,1b) for integers a≥1 and b≥0.
The trivial S -character is indexed by (n). Thus, χ(n)(µ) = 1 for all µ ⊢ n.
n
The next three lemmata collect a few more values of certain irreducible characters
that we shall need in the sequel. All statements are readily verified using e.g. the
Murnaghan-Nakayamarule.
Define fλ = χλ((1n)). This is the dimension of the irreducible representation
ρλ.
Lemma 2.1. Hook shape characters satisfy
n−1
f(n−k,1k) =
k
(cid:18) (cid:19)
and
χ(n−k,1k)((n))=(−1)k,
whereas χλ((n))=0 if λ is not a hook shape.
Lemma 2.2. Let λ⊢n. If λ has p parts of size 1 and q parts of size 2, then
χ(n−1,1)(λ)=p−1
STATISTICS OF RANDOM PRODUCTS 3
and
p−1
χ(n−2,12)(λ)= −q.
2
(cid:18) (cid:19)
A frequently occurring quantity is the content of λ⊢n. It is defined by
n χλ((2,1n−2))
c = .
λ 2 fλ
(cid:18) (cid:19)
Lemma 2.3. The content of a hook shape is
n(n−2k−1)
c = .
(n−k,1k) 2
3. Mean statistics and irreducible characters
Let s: S →R be any real-valued function on the symmetric group. In all our
n
subsequentapplications,swillbeapermutationstatisticassociatinganonnegative
integer with each permutation in S .
n
Choose Γ ⊆ S . We now describe a general procedure for computing the ex-
n
pected value of s on a product of random elements of Γ taken independently from
the uniform distribution.
The mean statistic s is the class function which computes the mean of s over
conjugacy classes. That is, s:P →R is defined by
n
1
s(λ)= s(π).
|C |
λ
πX∈Cλ
Hence,thinking ofclassfunctions asbeing definedonS , s=sif andonly ifs isa
n
class function. Considered in that particular setting, the remainder of this section
resembles the procedure described in [6].1
Let π ∈ S . We need to keep track of the ways to express π as a product of
n
elements from Γ. To this end, define a permutation statistic n by
t
n (π)=#{(γ ,...,γ )∈Γt |γ ···γ =π}.
t 1 t 1 t
Observethatalln areclassfunctionsifandonlyifΓisaunionofconjugacyclasses.
t
Theorem 3.1. If at least one of the statistics n and s is a class function, then
t
n!
E (s,t)= hs,n i.
Γ |Γ|t t
Proof. By definition,
1
E (s,t)= n (π)s(π).
Γ |Γ|t t
πX∈Sn
Assume now that s is a class function. Rewriting the right hand side by first
summing over the conjugacy classes of S , we obtain
n
1
E (s,t)= s(λ) n (π)
Γ |Γ|t t
λX⊢n πX∈Cλ
1
= s(λ)|C |n (λ)
|Γ|t λ t
λ⊢n
X
n!
= hs,n i,
|Γ|t t
as desired. If, instead, n is a classfunction, the proofis completely analogous. (cid:3)
t
1Thecontext of[6]was thatofs(π) beingtheabsolute lengthofπ withΓthesetoftranspo-
sitions. Thetechnique, however,couldclearlyhavebeenappliedtoanyclassfunctions.
4 AXELHULTMAN
Under the hypotheses of the preceding theorem, we are left with the task of
evaluatingthe innerproductoftwomeanstatistics. Thisiseasyifwearesomehow
able to express them in the orthonormal basis comprised of the irreducible S -
n
characters. In other words, we want to find the coefficients a and b(t) defined
λ λ
by
s= a χλ
λ
λ⊢n
X
and
n = b(t)χλ,
t λ
λ⊢n
X
respectively.
Although Theorem 3.1 applies if n or s is a class function, all our subsequent
t
applications come from the former setting. The next theorem is the reason; it
shows how to compute the b(t) if Γ consists of conjugacy classes. Variations of the
λ
formula (and its proof) are numerous in the literature. With a bit of labour, it
can be extracted e.g. from [3] or [7]. When Γ is a conjugacy class, the statement
followsimmediatelyfrom[10,TheoremA.1.9]whichisattributedtoFrobenius. We
provide a self-contained proof for convenience.
Theorem3.2. SupposeΓisadisjoint unionofconjugacyclassesΓ ,...,Γ ⊆S .
1 k n
Let µ denote the cycle type of the permutations in Γ . Then,
i i
t
k
1
b(t) = |Γ |χλ(µ ) .
λ n!(fλ)t−1 i i
!
i=1
X
Proof. Let δ denote the Kronecker delta. By the Schur orthogonality relations,
·,·
1
fλχλ(µ)=δ =n (µ).
n! (1n),µ 0
λ⊢n
X
This proves the t=0 case of the asserted statement.
Suppose f is any class function and define a linear map P on Cl by declaring
Γ n
P (f)(π)= f(πγ−1)
Γ
γ∈Γ
X
for π ∈ S . By definition, n = Pt(n ). Thus, it suffices to show that χλ is an
n t Γ 0
eigenfunction of P with eigenvalue 1 k |Γ |χλ(µ ). To this end, we may by
Γ fλ i=1 i i
linearityassumewithoutlossofgeneralitythatΓisasingleconjugacyclass. Denote
P
its cycle type simply by µ.
Define M = ρλ(γ). Since Γ is a conjugacy class, M and ρλ(π) commute
λ γ∈Γ λ
for all π ∈ S . By Schur’s Lemma, M = h(λ)I, where I is the identity map and
n λ
h(λ)∈C. Now oPbserve that
P (χλ)(π)=trace ρλ(π)M =h(λ)χλ(π)
Γ λ
for π ∈S . Plugging in π =id yields |(cid:0)Γ|χλ(µ)=(cid:1)h(λ)fλ, proving the claim. (cid:3)
n
For Γ satisfying the hypothesis of Theorem 3.2, we conclude that the remaining
challengeistodecomposethemeanstatisticsasalinearcombinationofirreducible
S -characters. The upcoming three sections are essentially devoted to such com-
n
putations.
STATISTICS OF RANDOM PRODUCTS 5
4. Cycle numbers
For π ∈S and a positive integer k, let cyc (π) denote the number of elements
n k
that are contained in k-cycles in the disjoint cycle decomposition of π. Thus, π
contains cyc (π)/k k-cycles. Clearly, cyc is a class function so that cyc = cyc .
k k k k
Thefollowingresult,providingadecompositionofthisstatistic,isdue toAlonand
Kozma [1]. We take this opportunity to state a shorter, independent proof.2
Theorem 4.1 (Theorem 3 in [1]). Let k ∈ [n]. Regarded as a class function on
S , cyc decomposes as
n k
min(k,n−k)
cyc =χ(n)+ (−1)k−iχ(n−k,i,1k−i)
k
i=1
X
k−1
+ (−1)k−jχ(j,n−k+1,1k−j−1).
j=n−k+1
X
Proof. Ifµ⊢n−k isobtainedfromλ⊢nbyremovingapartofsizek,weofcourse
havecyc (λ)=cyc (µ)+k. Thus,the symmetricfunctionimageofcyc underthe
k k k
characteristic map is
p cyc (λ) p p (cyc (µ)+k) p
λ k = µ k k =p µ =p s .
k k n−k
z z k(cyc (µ)/k+1) z
λ⊢n λ µ⊢n−k µ k µ⊢n−k µ
λhaXsk-parts X X
Using [14, 7.72], we may write
p s = (−1)ht(λ/(n−k))s ,
k n−k λ
X
where the sum is over all partitions λ ⊢ n such that λ/(n−k) is a border strip,
and ht(λ/(n−k)) is one less than the number of rows in the strip. Applying the
inverse of the characteristic map, this is precisely the desired result. (cid:3)
Theorem4.1refinesresultsfrom[6]and[9]. Theformerworkessentiallyrevolved
around the total number of cycles, i.e. the statistic cyc /k, whereas the fixed
k k
point number cyc was studied in the latter.
1 P
5. Excedances
Recall that an excedance of π ∈ S is an index i ∈ [n] such that π(i) > i.
n
Similarly, i is a weak excedance if π(i) ≥ i. Let exc(π) and wexc(π) denote the
number ofexcedancesandweak excedances,respectively,of π. Clearly,neither exc
nor wexc is a class function.
Theorem 5.1. The mean statistics exc and wexc decompose as
n−1 1
exc= χ(n)− χ(n−1,1)
2 2
and
n+1 1
wexc= χ(n)+ χ(n−1,1),
2 2
respectively.
2Theproofemploysstandardterminologyfromthetheoryofsymmetricfunctions. Werefrain
fromreproducingthe definitions sincethey arenot usedelsewhereinthe paper. Everything can
befounde.g.in[14,Chapter7].
6 AXELHULTMAN
Proof. If iis nota fixedpointofπ, then iis anexcedanceofπ ifandonly ifπ(i) is
not an excedance of π−1. A fixed point is a weak excedance but not an excedance.
Hence,
1 n−p
exc(λ)= (exc(π)+exc(π−1))= ,
2#C 2
λ
πX∈Cλ
where p is the number of fixed points of any π ∈C , i.e. the number of parts that
λ
equal one in λ. Similarly,
n+p
wexc(λ)= .
2
The result now follows from Lemma 2.2. (cid:3)
6. Inversions, descents and the major index
Thissectiontreatsthemeanstatisticsassociatedwiththreecommonlyoccurring
permutation statistics. First, we recall their definitions.
Letπ ∈S . Adescentofπ isanindex i∈[n−1]suchthatπ(i)>π(i+1). The
n
number of descents of π is denoted by des(π), whereas the major index maj(π) is
the sum of all descents of π.
An index pair 1 ≤ i < j ≤ n forms an inversion of π if π(i) > π(j). Let inv(π)
be the number of inversions of π.
Inordertostudythesestatisticsallatonce,itisconvenienttodefinethequantity
I (i,j)=#{π ∈C |π(i)>π(j)}
λ λ
for λ⊢n and 1≤i<j ≤n.
Lemma 6.1. Suppose λ ⊢ n. Let p and q denote the number of 1-parts and the
number of 2-parts, respectively, in λ. Then,
#C 2q−p(p−1) 2(j−i−1)((n−p)(1−p)−2q)
λ
I (i,j)= 1+ + .
λ
2 n(n−1) n(n−1)(n−2)
(cid:18) (cid:19)
Proof. Fix λ⊢n and indices 1≤i<j ≤n. Consider the following subsets of C :
λ
T ={π∈C |π(i)=i and π(j)=j},
1 λ
T ={π∈C |π(i)=j and π(j)=i},
2 λ
T ={π∈C |π(i)=i and π(j)=k for some i<k <j},
3 λ
T ={π∈C |π(j)=j and π(i)=k for some i<k<j},
4 λ
T ={π∈C |π(i)=j and π(j)=k for some i<k<j},
5 λ
T ={π∈C |π(j)=i and π(i)=k for some i<k <j},
6 λ
T =C \(T ∪T ∪T ∪T ∪T ∪T ).
7 λ 1 2 3 4 5 6
Thus, C is the disjoint union C =T ⊎···⊎T .
λ λ 1 7
Let f : C → C be the involution π 7→ (ij)π(ij), where (ij) denotes the
i,j λ λ
transposition which interchanges i and j. Then, f restricts to an involution
i,j
T →T . This restriction has no fixed points since T ∪T is the fixed point set of
7 7 1 2
f . Moreover, for a permutation π ∈ T , (i,j) is an inversion if and only if it is
i,j 7
not an inversion of f (π).
i,j
Observing that (i,j) is an inversion for all π ∈ T ∪T ∪T , whereas it is a
2 5 6
non-inversionfor all π ∈T ∪T ∪T , we thus obtain
1 3 4
#T
7
I (i,j)=#T +#T +#T +
λ 2 5 6
2
#C −#T +#T −#T −#T +#T +#T
λ 1 2 3 4 5 6
= .
2
STATISTICS OF RANDOM PRODUCTS 7
Computing
p(p−1)#C
λ
#T = ,
1
n(n−1)
2q#C
λ
#T = ,
2
n(n−1)
p(n−p)(j−i−1)#C
λ
#T =#T = ,
3 4
n(n−1)(n−2)
(n−p−2q)(j−i−1)#C
λ
#T =#T =
5 6
n(n−1)(n−2)
yields the desired result. (cid:3)
Next,weexploitthefactthatseveralfamiliarpermutationstatisticsareobtained
by taking appropriate sums of I (i,j).
λ
Theorem 6.2. The meanstatistics associated with des, majandinv canbewritten
as the following linear combinations of irreducible characters:
n−1 1 1
des= χ(n)− χ(n−1,1)− χ(n−2,1,1),
2 n n
n n(n−1) 1 1
maj= des= χ(n)− χ(n−1,1)− χ(n−2,1,1),
2 4 2 2
n(n−1) n+1 1
inv= χ(n)− χ(n−1,1)− χ(n−2,1,1).
4 6 6
Proof. Let λ⊢n. Applying Lemma 6.1, we obtain the identities
n−1
I (i,i+1) n−1 q 1 p
λ
des(λ)= = + − ,
#C 2 n n 2
i=1 λ (cid:18) (cid:19)
X
n−1
iI (i,i+1) n(n−1) q 1 p
λ
maj(λ)= = + − ,
#C 4 2 2 2
i=1 λ (cid:18) (cid:19)
X
I (i,j) n(n−1) p(p−1) n(p−1) q
λ
inv(λ)= = − − + ,
#C 4 12 6 6
λ
1≤i<j≤n
X
where p and q are as in Lemma 6.1. That these equations are equivalent to the
asserted ones is readily shown using Lemma 2.2. (cid:3)
7. Some examples
IfΓisaunionofconjugacyclasses,wemaycombineTheorem3.1withTheorem
3.2 in order to explicitly compute E (s,t) for any of the permutation statistics s
Γ
which were studied in the previous sections. We conclude the paper with a few
sample computations of this kind.
Example 7.1. Suppose Γ =T ⊆ S is the set of transpositions. The contents c
n λ
considered in Lemma 2.3 make an appearance as Theorem 3.2 specializes to
t
1 n χλ((2,1n−2)) ctfλ
b(t) = 2 fλ = λ .
λ n! fλ n!
(cid:0) (cid:1) !
Combining this with the decompositions found in Theorem 5.1 and Theorem 6.2,
we may invoke Theorem 3.1 and show that a product of t random transpositions
8 AXELHULTMAN
has the following expected values of the (weak) excedance number, descent number,
major index and inversion number, respectively:
t
n−1 2
E (exc,t)= 1− 1− ,
T
2 n−1
(cid:18) (cid:19) !
n+1 n−1 2 t
E (wexc,t)= 1+ 1− ,
T
2 n+1 n−1
(cid:18) (cid:19) !
t t
n−1 2 2 n−2 4
E (des,t)= 1− 1− − 1− ,
T
2 n n−1 n n−1
(cid:18) (cid:19) (cid:18) (cid:19) !
t t
n(n−1) 2 2 n−2 4
E (maj,t)= 1− 1− − 1− ,
T
4 n n−1 n n−1
(cid:18) (cid:19) (cid:18) (cid:19) !
n(n−1) 2(n+1) 2 t n−2 4 t
E (inv,t)= 1− 1− − 1− .
T
4 3n n−1 3n n−1
(cid:18) (cid:19) (cid:18) (cid:19) !
The formula for E (inv,t) obtained in Example 7.1 was previously found by
T
Sjo¨strand [13, Theorem 5.1].
WehaverefrainedfromstatingtheexplicitformulaforE (cyc ,t)whichfollows
T k
fromTheorem4.1. Withk =1,itrecoversresultsonfixedpointsfrom[9],whereas
the sum over all k leads to expected cycle numbers that were computed in [6].
Also, note that E (cyc ,t)/n is nothing but the probability that a product of t
T n
random transpositions forms an n-cycle. From that probability, one easily derives
the number of factorisations of an n-cycle into t transpositions. Working out the
details, one recovers the formula counting such factorisations which appears in
Jackson [8].
Example 7.2. Let Γ = C ⊆ S be the set of n-cycles. Assuming now that
(n) n
t>0, Theorem 3.2 and Lemma 2.1 show that b(t) =0 unless we have a hook shape
λ
λ=(n−j,1j). Moreover,
t
1 (n−1)!(−1)j n−1
b(t) = .
(n−j,1j) n! n−1 j
j ! (cid:18) (cid:19)
Suppose k < n and consider the statis(cid:0)tic (cid:1)cyc which counts elements that be-
k
long to k-cycles. According to Theorem 4.1, exactly two terms (namely χ(n) and
(−1)k−1χ(n−k,1k))which correspondtohookshapes appear intheexpansion ofcyc .
k
Applying Theorem 3.1, we deduce that the expected number of k-cycles in a product
of t>0 random n-cycles in S is
n
1 1 (−1)k(t+1)−1
E (cyc ,t)= + .
k C(n) k k k n−1 t−1
k
Letus,finally,consideranexamplewhereΓdo(cid:0)esno(cid:1)tconsistofasingleconjugacy
class.
Example 7.3. Suppose Γ is any union of conjugacy classes in which every permu-
tation has exactly one fixed point. In this case, Theorem 3.2 in conjunction with
Lemma 2.2 shows that b(t) =0 for all t>0. Thus, by Theorem 3.1 and Theo-
(n−1,1)
rem 5.1, the expected number of excedances of a product of t > 0 elements from Γ
is
n−1
E (exc,t)= ,
Γ
2
STATISTICS OF RANDOM PRODUCTS 9
whereas for weak excedances one obtains
n+1
E (wexc,t)= ,
Γ
2
independently of t.
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DepartmentofMathematics,Linko¨ping University,SE-58183,Linko¨ping, Sweden.