Table Of ContentCHARACTERS AND INVERSIONS
IN THE SYMMETRIC GROUP
Anne de M(cid:19)edicis, Victor Reiner, Mark Shimozono
Department of Mathematics
University of Minnesota
Minneapolis, MN 55455
November 1993
Abstract
We consider sums of the form
(cid:31)(cid:21)=(cid:22)((cid:25)) qinv((cid:25))
(cid:25)X2Sn
where (cid:31)(cid:21)=(cid:22) is a skew character of the symmetric group and inv((cid:25)) is the number
of inversions of (cid:25). Our main result gives a lower bound on the number of factors
of 1+q and 1(cid:0)q which dividethe abovesum, and isshownto be sharpwhen (cid:21)=(cid:22)
is a hook partition shape.
R(cid:19)esum(cid:19)e
Nous consid(cid:19)eronsdes sommes de la forme
(cid:31)(cid:21)=(cid:22)((cid:25)) qinv((cid:25))
(cid:25)X2Sn
ou(cid:18) (cid:31)(cid:21)=(cid:22) estuncaract(cid:18)eregauchedu groupesym(cid:19)etriqueetinv((cid:25)) d(cid:19)esignelenombre
d’inversions de (cid:25). Notre r(cid:19)esultat principal donne une borne inf(cid:19)erieure pour le
nombredefacteurs (1+q)et (1(cid:0)q)quidivisentcessommes. Nousd(cid:19)emontronsen
particulier que cette borne est exacte lorsque (cid:21)=(cid:22) est une(cid:19)equerre.
TypesetbyAMS-TEX
1
2 ANNE DE ME(cid:19)DICIS, VICTOR REINER, MARK SHIMOZONO
Section 1. Introduction
In this paper, we consider sums over the symmetric group S of the form
n
(cid:31)(cid:21)=(cid:22)((cid:25))qinv((cid:25))
(cid:25)X2Sn
where (cid:31)(cid:21)=(cid:22) is a skew character of the symmetric group [JP] and inv((cid:25)) is the
number of inversions of (cid:25), i.e.
inv((cid:25))=#f(i;j):1(cid:20)i<j (cid:20)n; (cid:25)(cid:0)1(i)>(cid:25)(cid:0)1(j)g:
There are two motivations for considering such sums. Firstly, other than some
special results of Gessel [Ge] and Edelman [Ed], there is very little known about
the joint distribution of inversions and conjugacy class (i.e. cycle type) in S .
n
Thisisinstarkcontrasttothe jointdistributionsknownforinversionsanddescent
statistics [GaGe], and for cycle type and descent statistics [GR]. Thus it seems
reasonable to ask what can be said about the sum
f((cid:25))qinv((cid:25))
(cid:25)X2Sn
whenf issomeclassfunction,i.e. afunctionwhichisconstantonconjugacyclasses
in S , such as an irreducible character (cid:31)(cid:21).
n
Remark
As pointed out by the referee, there is a di(cid:11)erent interpretation one can attach to
these sums, namely that
(cid:31)(cid:21)=(cid:22)((cid:25))qinv((cid:25)) =hs ;F i
(cid:21)=(cid:22) n
(cid:25)X2Sn
where h;i is the usual Hall inner product on symmetric functions, s is the skew
(cid:21)=(cid:22)
Schur function correspondingto (cid:21)=(cid:22), and F is a certain symmetric function with
n
coe(cid:14)cients in Z[q] which encodes all the information about the distribution of
inversions and conjugacy class in S :
n
F := qinv((cid:25))p
n (cid:21)((cid:25))
(cid:25)X2Sn
where p ((cid:25)) is the power sum symmetric function corresponding to the cycle type
(cid:21)
of (cid:25) (see [Mac]).
Thesecondmotivationis byanalogyto the workof [DF,Re]. Hereit wasshown
that sums of the form
(cid:31)(w)qdes((cid:25))
(cid:25)2W
X
have high divisibility by linear polynomial factors, where W is a classical Weyl
group, (cid:31) is a one-dimensional character of W, and des is the descent statistic on
W.
Our main theorem is the following
CHARACTERS AND INVERSIONS IN THE SYMMETRIC GROUP 3
Theorem 1. Let (cid:21)=(cid:22) be a skew shape with longest row of length r and longest
column of length s. Then
(cid:31)(cid:21)=(cid:22)((cid:25)) qinv((cid:25))
(cid:25)X2Sn
is divisible by
(1+q)br=2c(1(cid:0)q)bs=2c:
This Theorem is proven in Section 2.
The divisibilities in the previous theorem are sometimes tight, as demonstrated
by
Theorem 2. If (cid:21) is a hook partition (r;1s(cid:0)1) then
(cid:31)(cid:21)((cid:25)) qinv((cid:25)) =(1(cid:0)q)bs=2cS(q)
(cid:25)X2Sn
=(1+q)br=2cR(q)
where S(q);R(q)2Z[q] satisfy
(n+1)!bs=2c! s is even
S(1)= (s+1)!
( n!bs=2c! s is odd
s!
and
(n+1)!br=2c! r is even
R((cid:0)1)= (r+1)!
( n!br=2c! r is odd
r!
This is proven in Section 2. It is also conjectured there that S(q) has non-
negative coe(cid:14)cients whenever r (cid:21) s. In Section 3, we look more closely at the
very special case s=2 where can provethis conjecture by an interesting bijection.
In this special case, the results relate to the joint distribution of (cid:12)xed points and
inversions overthe symmetric group.
Section 2. Proof of Theorem 1
Theorem 1 will follow by a sequence of straightforward reductions from the
following lemma, which appears to be the essence of the divisibilities appearing in
this context. Notation: for A(cid:18)[n], let S be the subgroup of permutations in S
A n
which only permute within the elements of A and (cid:12)x all elements of [n](cid:0)A.
Lemma 3. Fix a subset A (cid:18) [n] having cardinality r and (cid:12)x any permutation
(cid:27) 2S . Then
[n](cid:0)A
qinv((cid:25)(cid:27))
(cid:25)X2SA
is divisible by (1+q)br=2c
Proof:. Make a change of variable to p=1+q. We need to show pbr=2c divides
(p(cid:0)1)inv((cid:25)(cid:27)) = p#J((cid:0)1)inv((cid:25)(cid:27))(cid:0)#J
(cid:25)X2SA (cid:25)X2SA
J(cid:18)Inv((cid:25))
= ((cid:0)1)inv((cid:27))+#Jp#J ((cid:0)1)inv((cid:25))
J(cid:18)X([n2]) J(cid:18)(cid:25)XI2nSvA((cid:25))
4 ANNE DE ME(cid:19)DICIS, VICTOR REINER, MARK SHIMOZONO
Here Inv((cid:25)) denotes the inversion set of (cid:25), that is
Inv((cid:25))=f(i;j):1(cid:20)i<j (cid:20)n; (cid:25)(cid:0)1(i)>(cid:25)(cid:0)1(j)g
So it su(cid:14)ces to show that for any J (cid:18) [n] with #J <br=2c we have
2
(cid:16) (cid:17)
((cid:0)1)inv((cid:25)) =0
(cid:25)X2SA
J(cid:18)Inv((cid:25))
Since 2#J (cid:20) r(cid:0)2, there must exist a pair i;j 2 A which are not involved in any
of the pairsin J, andhence multiplication on the left bythe transposition(ij) isa
sign-reversing involution which shows all terms in the above sum cancel. (cid:3)
Remark
The preceding lemma bears some resemblance to results of Bjo(cid:127)rner and Wachs
[BW] on the distribution of inversions over certain subsets of S which they call
n
generalized quotients. In particular,for certain of these generalizedquotients, they
produce nice hook-formula factorizations for the generating function of inversions,
whichnaturallyhavehighdivisibilityby(1+q). Itwouldbedesirabletounderstand
this connection better.
Proof of Theorem 1. We wishtoshowthat if(cid:21)=(cid:22) isaskewshapewith longestrow
of length r and longest column of length s, then
(cid:31)(cid:21)=(cid:22)((cid:25)) qinv((cid:25))
(cid:25)X2Sn
is divisible by
(1+q)br=2c(1(cid:0)q)bs=2c:
Our strategy will be to reduce the theorem to a very special case and then apply
the Lemma.
First note that if ((cid:21)=(cid:22))t denotes the skew shape which is the transpose of (cid:21)=(cid:22),
then
(cid:31)((cid:21)=(cid:22))t((cid:25)) qinv((cid:25)) = (cid:31)(cid:21)=(cid:22)((cid:25)) sgn((cid:25)) qinv((cid:25))
(cid:25)X2Sn (cid:25)X2Sn
= (cid:31)(cid:21)=(cid:22)((cid:25)) ((cid:0)q)inv((cid:25))
(cid:25)X2Sn
using the fact that
(cid:31)((cid:21)=(cid:22)t)((cid:25))=(cid:31)(cid:21)=(cid:22)((cid:25)) sgn((cid:25))
Therefore it su(cid:14)ces to show only divisibility by (1+q)br=2c.
Using the Littlewood-Richardson rule [Sa], one knows that
(cid:31)(cid:21)=(cid:22)((cid:25))= (cid:31)(cid:23)((cid:25))
(cid:23)
X
CHARACTERS AND INVERSIONS IN THE SYMMETRIC GROUP 5
where the sum rangesovera multiset of partitions (cid:23) which all havelongest part of
size at least r. Then using the Jacobi-Trudi identity [Sa], one can write
(cid:31)(cid:23)((cid:25))= (cid:6)(cid:31)H((cid:26))((cid:25))
(cid:26)
X
Here (cid:26) runs over a set of partitions of n, and H((cid:26)) is the skew diagram having
disjoint horizontal rows of size (cid:26) for each i.
i
Furthermore,expanding the Jacobi-Trudideterminant along its top rowtells us
that each (cid:26) appearingin the abovesum will haveits largestpart (cid:26) of size at least
1
r. Since we are only trying to show divisibility by (1+q)br=2c, it therefore su(cid:14)ces
to prove the theorem in the special case where (cid:21)=(cid:22) = H((cid:26)) for some partition (cid:26)
with largest part of size r.
Thecharacter(cid:31)H((cid:26)) is easytowrite downexplicitly, asit isthe characterofthe
permutation representation of S acting on the left cosets of the Young subgroup
n
S which permutes the (cid:12)rst (cid:26) numbers among themselves, permutes the next (cid:26)
(cid:26) 1 2
among themselves, etc. So (cid:31)H((cid:26))((cid:25)) is the number of such left cosets (cid:12)xed by (cid:25).
Therefore,
(cid:31)H((cid:26))((cid:25)) qinv((cid:25)) = #fcosets (cid:28)S :(cid:25)(cid:28)S =(cid:28)S g qinv((cid:25))
(cid:26) (cid:26) (cid:26)
(cid:25)X2Sn (cid:25)X2Sn
= qinv((cid:25))
coseXts(cid:28)S(cid:26) (cid:25)(cid:28)(cid:25)SX2(cid:26)=Sn(cid:28)S(cid:26)
= qinv((cid:25)1(cid:1)(cid:1)(cid:1)(cid:25)k)
(A1X;:::;Ak) ((cid:25)1X;:::;(cid:25)k)
]Ai=[n];#Ai=(cid:26)i (cid:25)i2SAi
= qinv((cid:25)1(cid:27))
]A(iA=1[Xn;:]:;:#;AAki=)(cid:26)i ((cid:25)(cid:25)2iX;2::S:;A(cid:25)ik)(cid:25)1X2SA1
where (cid:27) =(cid:25) (cid:1)(cid:1)(cid:1)(cid:25) in the last summation. By the Lemma, each of these sums
2 k
qinv((cid:25)1(cid:27))
(cid:25)1X2SA1
are divisible by (1+q)br=2c, and hence the theorem is proven. (cid:3)
Remark
At least some of the divisibility asserted by Theorem 1 is immediate, namely that
fors(cid:21)2the sumisdivisible by(1(cid:0)q)1,andbythe transposesymmetry,forr (cid:21)2
thesumisdivisibleby(1+q)1. Wesimplynotethatbythe Littlewood-Richardson
rule,fors(cid:21)2,(cid:31)(cid:21)=(cid:22) doesnot containthe trivialcharacterinits irreducibledecom-
position, and hence by the orthogonalityrelation for characters, we have
(cid:31)(cid:21)=(cid:22)((cid:25)) qinv((cid:25)) = (cid:31)(cid:21)=(cid:22)((cid:25))(cid:1)1((cid:25))=0
" #
(cid:25)X2Sn q=1 (cid:25)X2Sn
This implies the sum is divisible by (1(cid:0)q)1.
Section 3. Proof of Theorem 2
6 ANNE DE ME(cid:19)DICIS, VICTOR REINER, MARK SHIMOZONO
We only give a sketch of the proof.
To begin, we note that our earlier comment about the e(cid:11)ect of transposing the
shape(cid:21)=(cid:22)onthesumimpliesthatweonlyneedtoprovethatS(1)hastheasserted
value.
We will proceed by descending induction on r, beginning with the base case
r =n. Note that in this case (cid:31)(r;1s(cid:0)1) is the trivial character and
qinv((cid:25)) =[n]! =[n] [n(cid:0)1] (cid:1)(cid:1)(cid:1)[2] [1]
q q q q q
(cid:25)X2Sn
where [n] = 1+q +q2 +:::+qn(cid:0)1. The theorem then immediately follows in
q
this case from the aboveexpression. Forthe inductive step, ourstrategywill be to
replace(cid:31)(r;1s(cid:0)1) with a characterwecan compute more explicitly. Let (1s(cid:0)1)(cid:8)(r)
denote the skew shape which has a column of size s(cid:0)1 disjoint from a row of size
r. By an easy case of the Littlewood-Richardson rule,
(cid:31)(1s(cid:0)1)(cid:8)(r) =(cid:31)(r;1s(cid:0)1)+(cid:31)(r+1;1s(cid:0)2)
and hence
(cid:31)(r;1s(cid:0)1) =(cid:31)(1s(cid:0)1)(cid:8)(r)(cid:0)(cid:31)(r+1;1s(cid:0)2)
Using the last equation, one can check that the inductive step is equivalent to
proving the following:
(cid:31)(1s(cid:0)1)(cid:8)(r)((cid:25)) qinv((cid:25)) (n+2)n!br=2c! r is even
(cid:25)2Sn = (r+1)!
"P (1+q)br=2c #q=(cid:0)1 ( n!brr=!2c! r is odd
Toprovethelastexpression,wemakethe changeofvariablep=1+q asbefore,
andwewillusethefactthatthecharacter(cid:31)(1s(cid:0)1)(cid:8)(r)istheinductionfromS (cid:2)S
s(cid:0)1 r
to S of the charactersgn(cid:10)1. We then have
n
(cid:31)(1s(cid:0)1)(cid:8)(r)((cid:25)) qinv((cid:25))
(cid:25)2Sn
"P (1+q)br=2c #
q=(cid:0)1
(cid:31)(1s(cid:0)1)(cid:8)(r)((cid:25)) (p(cid:0)1)inv((cid:25))
= (cid:25)2Sn
"P pbr=2c #
p=0
=2p(cid:0)br=2c sgn((cid:25)j )(p(cid:0)1)inv((cid:25))3
A
66 (cid:25)X2Sn A2X(s[(cid:0)n]1) 77
6 (cid:25)(A)=A 7p=0
4 5
=2p(cid:0)br=2c sgn((cid:25)j ) p#J((cid:0)1)inv((cid:25))(cid:0)#J3
A
66 (cid:25)X2Sn A2X(s[(cid:0)n]1) J(cid:18)X([n2]) 77
6 (cid:25)(A)=A J(cid:18)Inv((cid:25)) 7p=0
4 5
=((cid:0)1)br=2c sgn((cid:25)j )
[n](cid:0)A
A2(s[(cid:0)n]1X);J(cid:18)([n2]) (cid:25)(A)=A(cid:25)X;2JS(cid:18)nInv((cid:25))
#J=br=2c
CHARACTERS AND INVERSIONS IN THE SYMMETRIC GROUP 7
The proof then proceeds by (cid:12)nding a sequence of relatively simple sign-reversing
involutions which sieve the above sum down to a small set of terms, each having
sign ((cid:0)1)br=2c. and having the desired cardinality to prove the assertion. (cid:3)
The data suggests that even more is true in the hook case:
Conjecture 4. Let (cid:21) be a hook partition (r;1s(cid:0)1), and S(q) as before, i.e
(cid:31)(cid:21)((cid:25)) qinv((cid:25)) =(1(cid:0)q)bs=2cS(q)
(cid:25)X2Sn
If r (cid:21)s, then S(q) has non-negative coe(cid:14)cients.
Section 4. The de(cid:12)ning representation
When s = 2, the character (cid:31)(r;1s(cid:0)1) is essentially the de(cid:12)ning character of S ,
n
i.e. the permutation representation of S acting on [n]. This character has the
n
particularly simple expression
(cid:31)(n(cid:0)1;1)((cid:25))=#Fix((cid:25))(cid:0)1
where Fix((cid:25)) is the set of (cid:12)xed points or 1-cycles of (cid:25) i.e.
Fix((cid:25))=fk:1(cid:20)k (cid:20)n; (cid:25) =kg
k
Therefore our results in this case have interpretations in terms of the joint distri-
bution of (cid:12)xed points and inversions over the symmetric group, some of which are
interesting. For example, it is well-known and easy to see that \the average per-
mutationinS hasone(cid:12)xedpoint"and\theaveragepermutationinS has 1 n
n n 2 2
inversions". The following corollary asserts that the value of #Fix((cid:25))(cid:1)inv((cid:25)) for
(cid:0) (cid:1)
the \averagepermutation (cid:25) in S " is
n
(3n+1)(n(cid:0)2)
12
Corollary 5.
(3n+1)(n(cid:0)2)
#Fix((cid:25))(cid:1)inv((cid:25))= n!
12
(cid:25)X2Sn
Proof. Let f(q) denote the sum
(cid:31)(n(cid:0)1;1)((cid:25))qinv((cid:25)) = (#Fix((cid:25))(cid:0)1)qinv((cid:25))
(cid:25)X2Sn (cid:25)X2Sn
Then we have
lim f(q)
S(1)=
q !1 1(cid:0)q
(cid:18) (cid:19)
d
= f(q)
dq
(cid:20) (cid:21)q=1
= (#Fix((cid:25))(cid:0)1)(cid:1)inv((cid:25))
(cid:25)X2Sn
By Theorem 2 in the case s=2, we have S(1)=(n+1)!=6. Combining this with
the easy fact that
n n!
inv((cid:25))=
2 2
(cid:25)X2Sn (cid:16) (cid:17)
gives the result. (cid:3)
The preceding result can be considerably strengthened:
8 ANNE DE ME(cid:19)DICIS, VICTOR REINER, MARK SHIMOZONO
Proposition 6.
n
k(cid:0)1 n(cid:0)k
#Fix((cid:25)) qinv((cid:25)) = [k(cid:0)1]! [n(cid:0)k]! qj2+2j
q q
j j
(cid:25)X2Sn Xk=1 Xj(cid:21)0 (cid:20) (cid:21)q(cid:20) (cid:21)q
where
n [n]!
q
=
k q [k]!q[n(cid:0)k]!q
. h i
Proof.
n
#Fix((cid:25)) qinv((cid:25)) = qinv((cid:25))
(cid:25)X2Sn Xk=1 (cid:25)(cid:25)X2k=Snk
n
= qinv((cid:27)) qinv((cid:28))
Xk=1 (cid:27)2S[1;k(cid:0)1](cid:2)XSfkg(cid:2)S[k+1;n] (cid:28)1<(cid:1)(cid:1)(cid:1)(cid:28)(cid:28)k2(cid:0)SX1n;;(cid:28)(cid:28)kk+=1k<(cid:1)(cid:1)(cid:1)(cid:28)n
where the last equality comes from the well-known fact that any permutation (cid:25)
may be factored uniquely as (cid:25) =(cid:27)(cid:28) with (cid:27) 2S (cid:2)S (cid:2)S and
[1;k(cid:0)1] fkg [k+1;n]
(cid:28) <(cid:1)(cid:1)(cid:1)<(cid:28) and (cid:28) <(cid:1)(cid:1)(cid:1)(cid:28)
1 k(cid:0)1 k+1 n
Note that we are restricting this factorization to the set of (cid:25) which (cid:12)x k, but this
creates no di(cid:14)culties. Since qinv((cid:25)) =[n]! , we have
(cid:25)2Sn q
P n
#Fix((cid:25)) qinv((cid:25)) = [k(cid:0)1]! [n(cid:0)k]! qinv((cid:25))
q q
(cid:25)X2Sn kX=1 (cid:28)1<(cid:1)(cid:1)(cid:1)(cid:28)(cid:28)k2(cid:0)SX1n;;(cid:28)(cid:28)kk+=1k<(cid:1)(cid:1)(cid:1)(cid:28)n
and it only remains to compute the last sum on the right-hand side. Given such a
(cid:28), de(cid:12)ne two subsets J ;J (cid:18)[n] by
1 2
J =fi:1(cid:20)i(cid:20)k(cid:0)1; (cid:28)(cid:0)1(i)(cid:21)k+1g
1
J =fi:k+1(cid:20)i(cid:20)n; (cid:28)(cid:0)1(i)(cid:20)k(cid:0)1g
2
NotethatJ ;J musthavethesamecardinalitywhichwedenotebyj,andchoosing
1 2
thesetsJ ;J completelydetermines(cid:28). Thereareexactlyj2 inversionsbetweenJ
1 2 1
andJ ,andexactly2j inversionsbetweenkandJ ;J . Thisaccountsfortheqj2+2j
2 1 2
term in the sum. The remaining inversions of (cid:28) come from among the numbers in
[1;k (cid:0)1] and among the numbers in [k +1;n], and correspond to MacMahon’s
\inv" statistic on subsets [Ma] applied to the sets J ;J respectively. Since the
1 2
distributionof\inv"overk-subsetsofann-setisgivenby n ,thetheoremfollows.
k q
(cid:3)
(cid:2) (cid:3)
Remark
The last sum on the right-hand side in the preceding theorem is similar to the
q-Vandermonde convolution
k(cid:0)1 n(cid:0)k n(cid:0)1
qj2 =
j j k
j(cid:21)0 (cid:20) (cid:21)q(cid:20) (cid:21)q (cid:20) (cid:21)q
X
CHARACTERS AND INVERSIONS IN THE SYMMETRIC GROUP 9
but does not sum so nicely. However, when we set q = 1 or q = (cid:0)1, the two are
the same.
Finally, we wish to prove the s = 2 case of Conjecture 4. Unravelling the
statement in this case, it conjectures that for all k (cid:20)n, one has
#Fix((cid:25))(cid:21)#f(cid:25) 2S :inv((cid:25))(cid:20)kg
n
in(cid:25)vX2((cid:25)S)n(cid:20)k
Recalling that the average permutation in S has one (cid:12)xed point, this may be
n
paraphrasedas sayingthat the set of all permutations with length at most k have,
on average, \more than their fair share" of (cid:12)xed points. This seems plausible,
sinceonewouldthinkthathavingfewinversionsandhaving(cid:12)xedpointsshouldbe
positively correlated. It is not hard to see that the assertion follows from
Theorem 7. There exists a bijection (cid:30) between the sets
f((cid:25);k):(cid:25) 2S ; k 2[n]; (cid:25) =kg$S
n k n
with the property that if (cid:30)((cid:25);k)=(cid:27) then inv((cid:27))(cid:21)inv((cid:25)).
Proof. First de(cid:12)ne the folding bijection f :[n]![n] by
2 minfk(cid:0)1;n(cid:0)kg+1 if k (cid:20)dne
f(k)= 2
2 minfk(cid:0)1;n(cid:0)kg+2 if k >dne
(cid:26) 2
Given a pair ((cid:25);k) as in the left-hand side of the theorem, de(cid:12)ne the permutation
(cid:27) = (cid:27) (cid:27) (cid:1)(cid:1)(cid:1)(cid:27) as follows: (cid:27) = f(k), and (cid:27) ;::: ;(cid:27) are the numbers [n](cid:0)f(k)
1 2 n 1 2 n
listed in the same relative order of magnitude as (cid:25) ;::: ;(cid:25) ;(cid:25) ;:::(cid:25) . It is
1 k(cid:0)1 k+1 n
easy to check that this is a bijection with the desired property. (cid:3)
Cansuchabijection(cid:30)befoundsothat(cid:27) (cid:21)(cid:25) inweak Bruhatorder? Thiswould
imply that the permutations in any lower order ideal of weak Bruhat order have,
on average,\more than their fair share" of (cid:12)xed points.
Section 5. Acknowledgements
The authors would like to thank Paul Edelman and Dennis Stanton for helpful
discussions and references.
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