Counting Permutations with Even Valleys and Odd … · Counting Permutations with Even Valleys and...
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Counting Permutations with Even Valleys and Odd Peaks Ira M. Gessel Department of Mathematics Brandeis University IMA Workshop Geometric and Enumerative Combinatorics University of Minnesota, Twin Cities November 13, 2014
Transcript of Counting Permutations with Even Valleys and Odd … · Counting Permutations with Even Valleys and...
Counting Permutations with Even Valleys andOdd Peaks
Ira M. Gessel
Department of MathematicsBrandeis University
IMA WorkshopGeometric and Enumerative Combinatorics
University of Minnesota, Twin CitiesNovember 13, 2014
Based on the paper
Counting permutations by alternating descents, by Ira M.Gessel and Yan Zhuang, Electronic Journal of Combinatorics21 (4), 2014, Paper #P4.23
Introduction
Let ⇡ = ⇡1⇡2 · · ·⇡n
be a permutation of [n] = {1, 2, . . . , n}. Theni is a peak if
⇡i�1 < ⇡
i
> ⇡i+1
j is a valley of ⇡ if⇡
j�1 > ⇡j
< ⇡j+1.
Example (peaks in blue, valleys in red):✓
11
24
35
42
56
67
73
◆
Introduction
Let ⇡ = ⇡1⇡2 · · ·⇡n
be a permutation of [n] = {1, 2, . . . , n}. Theni is a peak if
⇡i�1 < ⇡
i
> ⇡i+1
j is a valley of ⇡ if⇡
j�1 > ⇡j
< ⇡j+1.
Example (peaks in blue, valleys in red):✓
11
24
35
42
56
67
73
◆
On Math Overflow Liviu Nicolaescu, motivated by discreteMorse functions, asked the following question:
How many permutations of [n] are there are in which everyvalley is even and every peak is odd?
Let f (n) be the number of such permutations of [n]. We cancompute the first values by brute force:
n 0 1 2 3 4 5 6 7 8 9f(n) 1 1 2 4 13 50 229 1238 7614 52706
For example, with n = 3 the 4 permutations are
123, 213, 312, 321
(all except 132 and 231)
On Math Overflow Liviu Nicolaescu, motivated by discreteMorse functions, asked the following question:
How many permutations of [n] are there are in which everyvalley is even and every peak is odd?
Let f (n) be the number of such permutations of [n]. We cancompute the first values by brute force:
n 0 1 2 3 4 5 6 7 8 9f(n) 1 1 2 4 13 50 229 1238 7614 52706
For example, with n = 3 the 4 permutations are
123, 213, 312, 321
(all except 132 and 231)
On Math Overflow Liviu Nicolaescu, motivated by discreteMorse functions, asked the following question:
How many permutations of [n] are there are in which everyvalley is even and every peak is odd?
Let f (n) be the number of such permutations of [n]. We cancompute the first values by brute force:
n 0 1 2 3 4 5 6 7 8 9f(n) 1 1 2 4 13 50 229 1238 7614 52706
For example, with n = 3 the 4 permutations are
123, 213, 312, 321
(all except 132 and 231)
Sometimes exponential generating functions for certain classesof permutations defined by restrictions on descents have nicereciprocals. So let’s look at the reciprocal of
P1n=0 f (n)xn/n!.
We find that✓ 1X
n=0
f (n)xn/n!
◆�1
= 1 � x + 2x
3
3!� 5
x
4
4!+ 61
x
6
6!� 272
x
7
7!+ 7936
x
9
9!� · · ·
The coefficients are Euler numbers:1X
n=0
E
n
x
n
n!= sec x + tan x
= 1 + x +x
2
2!+ 2
x
3
3!+ 5
x
4
4!+ 16
x
5
5!+ 61
x
6
6!
+ 272x
7
7!+ 1385
x
8
8!+ · · ·
Sometimes exponential generating functions for certain classesof permutations defined by restrictions on descents have nicereciprocals. So let’s look at the reciprocal of
P1n=0 f (n)xn/n!.
We find that✓ 1X
n=0
f (n)xn/n!
◆�1
= 1 � x + 2x
3
3!� 5
x
4
4!+ 61
x
6
6!� 272
x
7
7!+ 7936
x
9
9!� · · ·
The coefficients are Euler numbers:1X
n=0
E
n
x
n
n!= sec x + tan x
= 1 + x +x
2
2!+ 2
x
3
3!+ 5
x
4
4!+ 16
x
5
5!+ 61
x
6
6!
+ 272x
7
7!+ 1385
x
8
8!+ · · ·
So it seems that
1X
n=0
f (n)x
n
n!=
✓1 � E1x + E3
x
3
3!� E4
x
4
4!+ E6
x
6
6!� E7
x
7
7!+ · · ·
◆�1
.
This is reminiscent of the generating function
✓1 � x +
x
3
3!� x
4
4!+
x
6
6!� x
7
7!+ · · ·
◆�1
for permutations with no increasing runs of length 3 or more,due to David and Barton (1962).
Is this just a coincidence?
So it seems that
1X
n=0
f (n)x
n
n!=
✓1 � E1x + E3
x
3
3!� E4
x
4
4!+ E6
x
6
6!� E7
x
7
7!+ · · ·
◆�1
.
This is reminiscent of the generating function
✓1 � x +
x
3
3!� x
4
4!+
x
6
6!� x
7
7!+ · · ·
◆�1
for permutations with no increasing runs of length 3 or more,due to David and Barton (1962).
Is this just a coincidence?
So it seems that
1X
n=0
f (n)x
n
n!=
✓1 � E1x + E3
x
3
3!� E4
x
4
4!+ E6
x
6
6!� E7
x
7
7!+ · · ·
◆�1
.
This is reminiscent of the generating function
✓1 � x +
x
3
3!� x
4
4!+
x
6
6!� x
7
7!+ · · ·
◆�1
for permutations with no increasing runs of length 3 or more,due to David and Barton (1962).
Is this just a coincidence?
Another form of the generating function is
✓1 � E1x + E3
x
3
3!� E4
x
4
4!+ E6
x
6
6!� E7
x
7
7!+ · · ·
◆�1
=3 sin
�12x
�+ 3 cosh
⇣12
p3x
⌘
3 cos�1
2x
��p
3 sinh⇣
12
p3x
⌘ .
Descents of permutations
Let ⇡ = ⇡1 · · ·⇡n
be a permutation of [n]. A descent of ⇡ is an i
such that ⇡i
> ⇡i+1. The descents of a permutation split it into
increasing runs, which are maximal consecutive increasingsubsequences.
For example the permutation 14528736 has descent set{3, 5, 6}:
1 4 5 · 2 8 · 7 · 3 6.
The increasing runs have lengths 3, 2, 1, 2.
We call the sequence of increasing run lengths the descentcomposition of ⇡. (It is a composition of n.) The descentcomposition has the same information as the descent set.
We’ll write D(L) for the descent set corresponding to L.
Descents of permutations
Let ⇡ = ⇡1 · · ·⇡n
be a permutation of [n]. A descent of ⇡ is an i
such that ⇡i
> ⇡i+1. The descents of a permutation split it into
increasing runs, which are maximal consecutive increasingsubsequences.
For example the permutation 14528736 has descent set{3, 5, 6}:
1 4 5 · 2 8 · 7 · 3 6.
The increasing runs have lengths 3, 2, 1, 2.
We call the sequence of increasing run lengths the descentcomposition of ⇡. (It is a composition of n.) The descentcomposition has the same information as the descent set.
We’ll write D(L) for the descent set corresponding to L.
Descents of permutations
Let ⇡ = ⇡1 · · ·⇡n
be a permutation of [n]. A descent of ⇡ is an i
such that ⇡i
> ⇡i+1. The descents of a permutation split it into
increasing runs, which are maximal consecutive increasingsubsequences.
For example the permutation 14528736 has descent set{3, 5, 6}:
1 4 5 · 2 8 · 7 · 3 6.
The increasing runs have lengths 3, 2, 1, 2.
We call the sequence of increasing run lengths the descentcomposition of ⇡. (It is a composition of n.) The descentcomposition has the same information as the descent set.
We’ll write D(L) for the descent set corresponding to L.
Descents of permutations
Let ⇡ = ⇡1 · · ·⇡n
be a permutation of [n]. A descent of ⇡ is an i
such that ⇡i
> ⇡i+1. The descents of a permutation split it into
increasing runs, which are maximal consecutive increasingsubsequences.
For example the permutation 14528736 has descent set{3, 5, 6}:
1 4 5 · 2 8 · 7 · 3 6.
The increasing runs have lengths 3, 2, 1, 2.
We call the sequence of increasing run lengths the descentcomposition of ⇡. (It is a composition of n.) The descentcomposition has the same information as the descent set.
We’ll write D(L) for the descent set corresponding to L.
Let L = (L1, . . . , Lk
) be a composition of n. Then we write�
n
L
�
for the multinomial coefficient
n!
L1! · · · Lk
!
Lemma. The number of permutations of n with descent setcontained in D(L) is
�n
L
�.
Proof by example. Take L = (3, 1, 2), so n = 6 andD(L) = {3, 4}. We take 1,2,. . . , 6 and put them into 3 boxeswith three in the first, one in the second, and two in the third:
| 2 4 5 | 1 | 3 6 |
We arrange the numbers in each box in increasing order andthen remove the bars to get
2 4 5 1 3 6
which has descent set contained in {3, 4}.
Let L = (L1, . . . , Lk
) be a composition of n. Then we write�
n
L
�
for the multinomial coefficient
n!
L1! · · · Lk
!
Lemma. The number of permutations of n with descent setcontained in D(L) is
�n
L
�.
Proof by example. Take L = (3, 1, 2), so n = 6 andD(L) = {3, 4}. We take 1,2,. . . , 6 and put them into 3 boxeswith three in the first, one in the second, and two in the third:
| 2 4 5 | 1 | 3 6 |
We arrange the numbers in each box in increasing order andthen remove the bars to get
2 4 5 1 3 6
which has descent set contained in {3, 4}.
Let L = (L1, . . . , Lk
) be a composition of n. Then we write�
n
L
�
for the multinomial coefficient
n!
L1! · · · Lk
!
Lemma. The number of permutations of n with descent setcontained in D(L) is
�n
L
�.
Proof by example. Take L = (3, 1, 2), so n = 6 andD(L) = {3, 4}. We take 1,2,. . . , 6 and put them into 3 boxeswith three in the first, one in the second, and two in the third:
| 2 4 5 | 1 | 3 6 |
We arrange the numbers in each box in increasing order andthen remove the bars to get
2 4 5 1 3 6
which has descent set contained in {3, 4}.
We can find the number of permutations of n with a givendescent set by inclusion-exclusion. Let us partial ordercompositions of n by reverse refinement, corresponding toordering descent sets by inclusion. So the compositions of 3are ordered as
(3)
(2,1)(1, 2)
(1,1,1)
Let �(L) be the number of permutations of [n] with descentcomposition L, where L is a composition of n.
Then�
n
L
�=
PKL
�(K ), so by inclusion-exclusion
�(L) =X
KL
(�1)l(L)�l(K )
✓n
K
◆,
where l(L) is the number of parts of L.
We can count all sorts of sets of permutations defined bydescent sets by added up �(L) for appropriate L. We getexponential generating functions since
�n
L
�is the coefficient of
x
n/n! inx
L
L!:=
x
L1
L1!· · · x
L
k
L
k
!.
Let �(L) be the number of permutations of [n] with descentcomposition L, where L is a composition of n.Then
�n
L
�=
PKL
�(K ), so by inclusion-exclusion
�(L) =X
KL
(�1)l(L)�l(K )
✓n
K
◆,
where l(L) is the number of parts of L.
We can count all sorts of sets of permutations defined bydescent sets by added up �(L) for appropriate L. We getexponential generating functions since
�n
L
�is the coefficient of
x
n/n! inx
L
L!:=
x
L1
L1!· · · x
L
k
L
k
!.
Let �(L) be the number of permutations of [n] with descentcomposition L, where L is a composition of n.Then
�n
L
�=
PKL
�(K ), so by inclusion-exclusion
�(L) =X
KL
(�1)l(L)�l(K )
✓n
K
◆,
where l(L) is the number of parts of L.
We can count all sorts of sets of permutations defined bydescent sets by added up �(L) for appropriate L. We getexponential generating functions since
�n
L
�is the coefficient of
x
n/n! inx
L
L!:=
x
L1
L1!· · · x
L
k
L
k
!.
Alternating descents
Following Denis Chebikin (2008), we define an alternatingdescent of a permutation ⇡ to be an odd descent or an evenascent. For example the alternating descents of 3 7 4 2 1 5 6 8are 3 and 6:
3 7 4 · 2 1 5 · 6 8
The alternating descents of a permutation split it up intoalternating runs. Each alternating run is either an “up-down"permutation if it starts in an odd position or a “down-up"permutation if it starts in an even position.
Nicolaescu’s question is closely connected to alternating runsof permutations, because a permutation has all valleys evenand all peaks odd if and only if all of its alternating runs havelength less than 3.
Alternating descents
Following Denis Chebikin (2008), we define an alternatingdescent of a permutation ⇡ to be an odd descent or an evenascent. For example the alternating descents of 3 7 4 2 1 5 6 8are 3 and 6:
3 7 4 · 2 1 5 · 6 8
The alternating descents of a permutation split it up intoalternating runs. Each alternating run is either an “up-down"permutation if it starts in an odd position or a “down-up"permutation if it starts in an even position.
Nicolaescu’s question is closely connected to alternating runsof permutations, because a permutation has all valleys evenand all peaks odd if and only if all of its alternating runs havelength less than 3.
Alternating descents
Following Denis Chebikin (2008), we define an alternatingdescent of a permutation ⇡ to be an odd descent or an evenascent. For example the alternating descents of 3 7 4 2 1 5 6 8are 3 and 6:
3 7 4 · 2 1 5 · 6 8
The alternating descents of a permutation split it up intoalternating runs. Each alternating run is either an “up-down"permutation if it starts in an odd position or a “down-up"permutation if it starts in an even position.
Nicolaescu’s question is closely connected to alternating runsof permutations, because a permutation has all valleys evenand all peaks odd if and only if all of its alternating runs havelength less than 3.
This is because an odd valley or an even peak corresponds toan alternating run of length at least 3:
✓11
26
32. . .
65
73
84. . .
◆
As Chebikin observed, the number of permutations of [n] withalternating descent set contained in D(L) is
✓n
L
◆
E
:=n!
L1! · · · Lk
!E
L1 · · ·EL
k
We prove this by putting the numbers 1,. . . , n into boxes with L
i
in the i th box, and then arranging the numbers in each box intoeither an up-down or down-up permutation, depending on theparity of the starting position.
Now let �(L) be the number of permutations of [n] withalternating descent composition L, where L is a composition ofn. Then just as before,
�n
L
�E
=P
KL
�(K ), so byinclusion-exclusion
�(L) =X
KL
(�1)l(L)�l(K )
✓n
K
◆
E
.
Note that�
n
L
�E
is the coefficient of x
n/n! in
E
L1
x
L1
L1!· · ·E
L
k
x
L
k
L
k
!.
Now let �(L) be the number of permutations of [n] withalternating descent composition L, where L is a composition ofn. Then just as before,
�n
L
�E
=P
KL
�(K ), so byinclusion-exclusion
�(L) =X
KL
(�1)l(L)�l(K )
✓n
K
◆
E
.
Note that�
n
L
�E
is the coefficient of x
n/n! in
E
L1
x
L1
L1!· · ·E
L
k
x
L
k
L
k
!.
Therefore, if we can find an exponential generating function forsome class of permutations determined by descent sets,obtained by adding up �(L)xL/L! for appropriate L, then bychanging x
n/n! to E
n
x
n/n! we should get the correspondingresult for alternating descent sets. This will explain theconnection between
✓1 � x +
x
3
3!� x
4
4!+
x
6
6!� x
7
7!+ · · ·
◆�1
and✓
1 � E1x + E3x
3
3!� E4
x
4
4!+ E6
x
6
6!� E7
x
7
7!+ · · ·
◆�1
.
Unfortunately, just changing to x
n/n! with E
n
x
n/n! doesn’treally work:
we would like a linear map that takes
x
L1
L1!· · · x
L
k
L
k
!.
to
E
L1
x
L1
L1!· · ·E
L
k
x
L
k
L
k
!.
but we can’t do this since these products aren’t linearlyindependent.
Unfortunately, just changing to x
n/n! with E
n
x
n/n! doesn’treally work: we would like a linear map that takes
x
L1
L1!· · · x
L
k
L
k
!.
to
E
L1
x
L1
L1!· · ·E
L
k
x
L
k
L
k
!.
but we can’t do this since these products aren’t linearlyindependent.
For example,x
2
2!· x
3
3!= 10
x
5
5!,
but
E2x
2
2!· E3
x
3
3!6= 10E5
x
5
5!
To make this work, and to compute the generating functionsmore efficiently, we use noncommutatative symmetricfunctions, which were studied extensively by Gelfand, Krob,Lascoux, Leclerc, Retakh, and Thibon in 1995 and furtherstudied by Thibon and others since then. But we will need onlysome very simple properties of the noncommutative symmetricfunctions most of which were known earlier.
We work with formal power series in noncommuting variablesX1,X2, . . . . We define the complete noncommutative symmetricfunctions
hn
=X
i1···i
n
X
i1X
i2 · · ·Xi
n
.
For any composition L = (L1, . . . , Lk
) of n we define
hL
= hL1 · · ·hL
k
which can be written as
hL
=X
L
X
i1X
i2 · · ·Xi
n
where the sum is over all (i1, . . . , in) satisfying
i1 · · · i
L1| {z }L1
, i
L1+1 · · · i
L1+L2| {z }L2
, · · · , i
L1+···+L
k�1+1 · · · i
n| {z }L
k
.
We define the ribbon noncommutative symmetric functions
rL
=X
L
X
i1X
i2 · · ·Xi
n
where the sum is over all (i1, . . . , in) satisfying
i1 · · · i
L1| {z }L1
> i
L1+1 · · · i
L1+L2| {z }L2
> · · · > i
L1+···+L
k�1+1 · · · i
n| {z }L
k
.
Thenh
L
=X
KL
rK
,
so by inclusion-exclusion,
rL
=X
KL
(�1)l(L)�l(K )hK
The algebra of noncommutative symmetric functions isgenerated by the h
L
.
When we make the X
i
commute, we get ordinary symmetricfunctions: h
L
becomes the ordinary complete symmetricfunction h
L
, and rL
becomes the ribbon (skew-hook, zigzag)skew Schur function.
The algebra of noncommutative symmetric functions isgenerated by the h
L
.
When we make the X
i
commute, we get ordinary symmetricfunctions: h
L
becomes the ordinary complete symmetricfunction h
L
, and rL
becomes the ribbon (skew-hook, zigzag)skew Schur function.
We can define a homomorphism from noncommutativesymmetric functions to exponential generating functions by
�(hn
) =x
n
n!
so
�(hL
) =x
L1
L1!· · · x
L
k
L
k
!=
✓n
L
◆x
n
n!.
Note that for a noncommutative symmetric function f , �(f ) isthe coefficient of x1x2 · · · xn
in the result of replacing X1,X2, . . .with commuting variables x1, x2, . . . .Then by inclusion-exclusion,
�(rL
) = �(L)x
n
n!.
Similarly, we can define another homomorphism fromnoncommutative symmetric functions to exponential generatingfunctions by
�(hn
) = E
n
x
n
n!
so by the inclusion-exclusion formula for � we have
�(rL
) = �(L)x
n
n!.
This means that if we can find the noncommutative symmetricfunction generating function for some class of wordsdetermined by their descent sets (i.e., we can express it interms of the r
L
), then by applying � and � we can get thecorresponding exponential generating functions forpermutations with the corresponding descent sets oralternating descent sets.
There are many formulas to which we can apply � and � , butone that gives us what we need for Nicolaescu’s problem is thefollowing:
This means that if we can find the noncommutative symmetricfunction generating function for some class of wordsdetermined by their descent sets (i.e., we can express it interms of the r
L
), then by applying � and � we can get thecorresponding exponential generating functions forpermutations with the corresponding descent sets oralternating descent sets.
There are many formulas to which we can apply � and � , butone that gives us what we need for Nicolaescu’s problem is thefollowing:
Theorem 1. (G, Jackson and Aleliunas; 1977) Let w1,w2, . . . bearbitrary commuting weights. Then
X
L
w
L
rL
=
✓ 1X
n=0
a
n
hn
◆�1
where the sum on the left is over all compositionsL = (L1, . . . , Lk
), and w
L
= w
L1 · · ·wL
k
.Here a0, a1, . . . are defined by
1X
n=0
a
n
z
n =
✓ 1X
n=0
w
n
z
n
◆�1,
where w0 = 1.
If we take w
i
= 1 for i < m and w
i
= 0 for i � m in Theorem 1then we get
X
L
rL
=
✓ 1X
n=0
(hmn
� hmn+1)
◆�1,
where the sum on the left is over compositions L with all partsless than m.
Applying � gives David and Barton’s result that the exponentialgenerating function for permutations in which every increasingrun has length less than m is
✓1 � x +
x
m
m!� x
m+1
(m + 1)!+
x
2m
(2m)!� x
2m+1
(2m + 1)!+ · · ·
◆�1
,
If we take w
i
= 1 for i < m and w
i
= 0 for i � m in Theorem 1then we get
X
L
rL
=
✓ 1X
n=0
(hmn
� hmn+1)
◆�1,
where the sum on the left is over compositions L with all partsless than m.
Applying � gives David and Barton’s result that the exponentialgenerating function for permutations in which every increasingrun has length less than m is
✓1 � x +
x
m
m!� x
m+1
(m + 1)!+
x
2m
(2m)!� x
2m+1
(2m + 1)!+ · · ·
◆�1
,
Applying � gives" 1X
n=0
✓E
mn
x
mn
(mn)!� E
mn+1x
mn+1
(mn + 1)!
◆#�1
as the exponential generating function for permutations inwhich every alternating run has length less than m; the casem = 3 is the solution to Nicolaescu’s problem.
Two more special cases of Theorem 1
If we set w
m
= 1 and w
n
= 0 for n 6= m then we find that
1X
n=0
r(mn) =
✓ 1X
n=0
(�1)n hmn
◆�1,
Applying � gives Carlitz’s 1973 result that✓ 1X
n=0
(�1)n
x
mn
(mn)!
◆�1
is the exponential generating function for permutations in whichevery increasing run has length m.
Applying � shows that✓ 1X
n=0
(�1)n
E
mn
x
mn
(mn)!
◆�1
counts permutations in which every alternating run haslength m.
For m = 4, these are permutations of [4n] with descent set{2, 6, 10, . . . , 4n � 2}.
Applying � shows that✓ 1X
n=0
(�1)n
E
mn
x
mn
(mn)!
◆�1
counts permutations in which every alternating run haslength m.
For m = 4, these are permutations of [4n] with descent set{2, 6, 10, . . . , 4n � 2}.
If we set w
n
= t for all n � 1 in Theorem 1, then we arecounting words by the number of increasing runs (which is onemore than the number of descents), and we get
X
L
t
l(L)rL
= (1 � t)
1 � t
1X
n=0
(1 � t)n hn
��1,
where the first sum is over all compositions L.
Applying � gives the well-known generating function for theEulerian polynomials,
1 +1X
n=1
A
n
(t)x
n
n!=
1 � t
1 � te
(1�t)x,
whereA
n
(t) =X
⇡2Sn
t
des(⇡)+1,
and des(⇡) is the number of descents of ⇡.
If we set w
n
= t for all n � 1 in Theorem 1, then we arecounting words by the number of increasing runs (which is onemore than the number of descents), and we get
X
L
t
l(L)rL
= (1 � t)
1 � t
1X
n=0
(1 � t)n hn
��1,
where the first sum is over all compositions L.
Applying � gives the well-known generating function for theEulerian polynomials,
1 +1X
n=1
A
n
(t)x
n
n!=
1 � t
1 � te
(1�t)x,
whereA
n
(t) =X
⇡2Sn
t
des(⇡)+1,
and des(⇡) is the number of descents of ⇡.
Applying � gives
1 +1X
n=1
A
n
(t)x
n
n!=
1 � t
1 � t
�sec(1 � t)x + tan(1 � t)x
� ,
whereA
n
(t) =X
⇡2Sn
t
altdes(⇡)+1
and altdes(⇡) is the number of alternating descents of ⇡, aresult due to Chebikin.
The major index and alternating major index
We define the major index maj(⇡) of a permutation ⇡ to be thesum of the descents of ⇡, and we define the alternating majorindex altmaj(⇡) to be the sum of the alternating descents of ⇡.
Then there is an analogue of Theorem 1 that allows us to countpermutations by major index and alternating major index.
I’ll just state the results:
1X
n=0
x
n
n!
P⇡2S
n
t
des(⇡)q
maj(⇡)
(1 � t)(1 � tq) · · · (1 � tq
n)=
1X
k=0
t
k
kY
j=0
e
xq
j
(MacMahon/Riordan 1954/Carlitz 1975)
1X
n=0
x
n
n!
P⇡2S
n
t
altdes(⇡)q
altmaj(⇡)
(1 � t)(1 � tq) · · · (1 � tq
n)
=1X
k=0
t
k
kY
j=0
�sec(xq
j) + tan(xq
j)�
(Remmel)
The remaining material was not presented during my talk.
More homomorphismsThere are other homomorphisms from the ring ofnoncommutative symmetric functions that enable us to countpermutations by other parameters.
The simplest one allows us to count permutations byinversions. An inversion of a permutation ⇡ is a pair (i , j) withi < j such that ⇡
i
> ⇡j
. We define a homomorphism �1 onnoncommutative symmetric function by
�1(hn
) =x
n
(q)n
,
where (q)n
= (1 � q)(1 � q
2) · · · (1 � q
n). Then if L is acomposition of n, we have
�1(rL
) =X
C(⇡)=L
q
inv(⇡) x
n
(q)n
.
More homomorphismsThere are other homomorphisms from the ring ofnoncommutative symmetric functions that enable us to countpermutations by other parameters.
The simplest one allows us to count permutations byinversions. An inversion of a permutation ⇡ is a pair (i , j) withi < j such that ⇡
i
> ⇡j
. We define a homomorphism �1 onnoncommutative symmetric function by
�1(hn
) =x
n
(q)n
,
where (q)n
= (1 � q)(1 � q
2) · · · (1 � q
n). Then if L is acomposition of n, we have
�1(rL
) =X
C(⇡)=L
q
inv(⇡) x
n
(q)n
.
We define the inverse major index imaj(⇡) to be maj(⇡�1).Then it’s also true that
�1(rL
) =X
C(⇡)=L
q
imaj(⇡) x
n
(q)n
.
Thus, for example,
✓1 � x +
x
3
(q)3� x
4
(q)4+
x
6
(q)6� x
7
(q)7+ · · ·
◆�1
is the Eulerian generating function for permutations with noincreasing runs of length 3 by inversions or by inverse majorindex.
We define the inverse major index imaj(⇡) to be maj(⇡�1).Then it’s also true that
�1(rL
) =X
C(⇡)=L
q
imaj(⇡) x
n
(q)n
.
Thus, for example,
✓1 � x +
x
3
(q)3� x
4
(q)4+
x
6
(q)6� x
7
(q)7+ · · ·
◆�1
is the Eulerian generating function for permutations with noincreasing runs of length 3 by inversions or by inverse majorindex.
More generally, if we make the variables X
i
commute, we getthe quasi-symmetric generating function for countingpermutations by their inverse descent set, from which we cancount by the number of inverse descents, or the number ofinverse peaks, etc.
These homomorphisms may be defined more naturally on alarger algebra than the noncommutative symmetric functions,the Malvenuto-Reutenauer algebra (also called the algebra offree quasi-symmetric functions).
To define it, we first define the standardization st(w) of the wordw to be the permutation obtained by replacing the smallestentry by 1, the next smallest by 2, and so on, where repeatedletters are ordered from left to right, so for example
st(1 1 5 2 1 2) = 1 2 6 4 3 5.
These homomorphisms may be defined more naturally on alarger algebra than the noncommutative symmetric functions,the Malvenuto-Reutenauer algebra (also called the algebra offree quasi-symmetric functions).
To define it, we first define the standardization st(w) of the wordw to be the permutation obtained by replacing the smallestentry by 1, the next smallest by 2, and so on, where repeatedletters are ordered from left to right, so for example
st(1 1 5 2 1 2) = 1 2 6 4 3 5.
Then for each ⇡ 2 Sn
we define a series [⇡] in thenoncommuting variables X1,X2, . . . by
[⇡] =X
st(i1···in)=⇡
X
i1 · · ·Xi
n
For example,[132] =
X
ab<c
X
a
X
c
X
b
and[12 · · · n] = h
n
.
The Malvenuto-Reutenauer algebra is spanned by the [⇡] with⇡ 2 S
n
for some n.
It’s easy to describe how these basis elements multiply: if⇡ 2 S
m
and � 2 Sn
then
[⇡][�] =X
[⌧ ]
where the sum is over all ⌧ = ⌧1 · · · ⌧m+n
2 Sm+n
for whichst(⌧1 · · · ⌧m
) = ⇡ and st(⌧m+1 · · · ⌧m+n
) = �.
The Malvenuto-Reutenauer algebra is spanned by the [⇡] with⇡ 2 S
n
for some n.
It’s easy to describe how these basis elements multiply: if⇡ 2 S
m
and � 2 Sn
then
[⇡][�] =X
[⌧ ]
where the sum is over all ⌧ = ⌧1 · · · ⌧m+n
2 Sm+n
for whichst(⌧1 · · · ⌧m
) = ⇡ and st(⌧m+1 · · · ⌧m+n
) = �.
Then the algebra of noncommutative symmetric functions is thesubalgebra of the Malvenuto-Reutenauer algebra generating bythe h
n
= [1 · · · n].
Then the homomorphism �1 extends in two different ways toMR, we can define either
�1a([⇡]) = q
inv(⇡) x
n
(q)n
or�1b([⇡]) = q
imaj(⇡) x
n
(q)n
,
where ⇡ 2 Sn
.
We have the following relations between the algebras thatwe’ve discussed:
NCSF ����! MR??y
??y
SF ����! QSF
