Journal of Algebraic Combinatorics 13 (2001), 83–101
c 2001 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
Vertex Operators for Standard Bases
of the Symmetric Functions
MIKE ZABROCKI
zabrocki@math.uqam.ca
CRM, Univeristé de Montréal, LaCIM, Université du Québec à Montréal, C.P. 8888, Succ. A, Montréal,
Canada H3C 3P8
Received April 27, 1999; Revised February 15, 2000
Abstract. We present formulas for operators which add a row or a column to the partition indexing the power,
monomial, forgotten, Schur, homogeneous and elementary symmetric functions. As an application of these operators we show that the operator that adds a column to the Schur functions can be used to calculate a formula for
the number of pairs of standard tableaux the same shape and height less than or equal to a fixed k.
Keywords: symmetric functions, vertex operators
1.
Notation
Using the notation of [3], we will consider the power { pλ [X ]}λ , Schur {sλ [X ]}λ , monomial
{m λ [X ]}λ , homogeneous {h λ [X ]}λ , elementary {eλ [X ]}λ and forgotten { f λ [X ]}λ bases for
the symmetric functions. We will often appeal to [3] for proofs any symmetric function
identities.
These bases are all indexed by partitions, non-increasing sequences of non- negative
integers. The ith entry of the partition will be denoted by λi . The length of a partition λ is
the largest i such that λi is non-zero and will be denoted by l(λ). The size of the partition
will be denoted by |λ| and is equal to the sum over all the entries of λ. The symbol n i (λ)
will be used to represent the number of parts of size i in the partition λ. The conjugate
partition will be denoted by λ0 and is the partition such that λi0 = the number of j such that
λ j is greater than or equal to i.
There is a standard inner product on
Qsymmetric functions h pλ , pµ i = z λ δλµ where δx y = 1
if x = y and 0 otherwise and z λ = i≥1 i ni (λ) n i (λ)!.
We will use a few non-standard operations on partitions that will require some notation.
The first is adding a column (or a sequence of columns) to a partition. Let a k | λ denote
the partition (λ1 + a, λ2 + a, . . . , λk + a). We will assume that this partition is undefined
when l(λ) > k.
Use the notation λ − (µ) to denote the partition formed by removing the parts that
are equal to µ from the partition λ. This of course assumes that there is a sequence I =
{i 1 , i 2 , . . . il(µ) } ⊂ {1, 2, . . . l(λ)} such that µ = (λi1 , λi2 , . . . , λil(µ) ). If this sequence does
not exist then λ − (µ) is again undefined.
84
ZABROCKI
The last operation will be inserting parts into a partition and will be represented by λ+(µ).
This will be the partition formed by ordering the sequence (λ1 , λ2 , . . . , λl(λ) , µ1 , µ2 , . . . ,
µl(µ) ) into a partition.
Occasionally within sums we will have an expression such as m λ−(µ) or h 1k |λ when it is
not necessarily the case that µ is a subpartition of λ or that λ has height less than or equal
k. We will consider expressions like these to be zero. This simplifies many summation
formulas which otherwise would have to have three or four conditions to their arguments,
but it is important to verify at each step that transformations to equations are in fact legal.
We will say that two bases for the symmetric functions {aλ }λ and {bλ }λ are dual if they
have the property that haλ , bµ i = δλµ . By definition, the power symmetric functions are dual
to the basis { pλ /z λ }λ . The monomial and homogeneous symmetric functions are dual. The
forgotten and the elementary symmetric functions are dual. The Schur symmetric functions
are self dual ( hsλ , sµ i = δλµ ).
There exists an involution, ω, on symmetric functions that relates the elementary and
homogeneous bases by ωh µ = eµ and the monomial and forgotten bases by ωm µ = f µ . It
also has the property that ωsλ = sλ0 .
Denote the operation of ‘skewing’ by a symmetric function f by f ⊥ . It is defined as
being the operation dual to multiplication by the symmetric function f in the sense that
h f ⊥ P, Qi = hP, f Qi. Its
P action on an arbitrary symmetric function P may be calculated
by the formula f ⊥ P = λ hP, f aλ ibλ for any dual bases {aλ }λ and {bλ }λ .
P
Using results and notation in [3] (p. 92–93 example (I.5.25)), by setting 1P
f = µ (aµ⊥ f )
{aµ }µ and {bµ }µ are any dual bases. It follows that if 1 f = i ci ⊗ di then
⊗ bµ where P
f ⊥ (P Q) = i ci⊥ (P)di⊥ (Q). Using this result and considering f to be multiplication by
some symmetric function we have,
X¡
¢
h i⊥ f h ⊥
k−i
(1)
¢ ⊥
ei⊥ f ek−i
(2)
¡
¢
pk⊥ f = pk⊥ f + f pk⊥
(3)
h⊥
k f =
i
ek⊥ f =
X¡
i
By the phrase ‘vertex operators’ we are referring to linear symmetric function operators
that add a row or a column to the partitions indexing a particular family of symmetric
functions. Formulas of this type for symmetric functions are sometimes called Rodrigues
formulas. In this article we look at those symmetric function operators which lie in the
linear span of { f i gi⊥ }i where f i and gi are symmetric functions to find expressions for
vertex operators for each basis.
By Corollary 3 in Section 4 we know that it is always possible to produce such a vertex
operator. Yet for some applications a more refined formula is necessary, and it seems that
only for a very small class of operators will this formula reduce to something elegant.
The vertex operators for the elementary and forgotten symmetric function basis are
related to the operators for the homogeneous and monomial (resp.) symmetric functions by
conjugating by the operator ω.
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
85
The existence of such operators for the Schur ([3] p. 96–97, [5] p. 69), and (row only)
Hall-Littlewood ([3] p. 237–238, [2]) symmetric functions are known. For the multiplicative
bases, it is clear that there exists operators that add a row to the symmetric functions of this
form since ek eλ = eλ+(k) , h k h λ = h λ+(k) and pk pλ = pλ+(k) , but adding a column is not an
obvious operation.
In general, formulas for adding a row or a column can be useful in proving a combinatorial interpretation for a symmetric function or deriving new formulas or properties. The
author’s interest in this particular question comes from trying to find vertex operators for
the Macdonald symmetric functions. The Macdonald vertex operator must specialize to the
vertex operators for other symmetric functions and so understanding these operators is an
important first step.
2.
The power vertex operator
This is the warm up case for the other 5 bases. The commutation relation between pk⊥ and
p j (given by Eq. (3)) has a nice expression: pk⊥ p j = p j pk⊥ + kδk j . This can be used to show
⊥
(where it is assumed that
the slightly more general relation pλ⊥ pk = pk pλ⊥ + kn k (λ) pλ−(k)
⊥
pλ−(k) = 0 if λ does not contain a part of size k).
The vertex operator is given by the following theorem
Theorem 1
CPa k =
For a ≥ 0 and k ≥ 0 define the following linear operator
X
λ : l(λ)≤k
pak−l(λ)
l(λ)
Y
±
( pλi +a − pλi pa ) pλ⊥ z λ
i=1
where the sum is over all partitions λ with less than or equal to k parts (if k = 0 then
CPa 0 = 1). CPa k has the property that CPa k pµ = pa k |µ for all µ such that l(µ) < k.
Proof: The proof is by induction on the number of parts of µ. Clearly this operator has
the property that CPa k 1 = pak since pλ⊥ kills 1 for |λ| > 0. From the commutation relation
of pλ⊥ and pk we derive that
CPa k p j = ( p j+a − p j pa )CPa k−1 + p j CPa k
(4)
The proof by induction follows from this relation.
2
The formula CPa k was chosen so that it has two properties: it adds a column to the power
symmetric functions, and it has a relatively simple expression when written in this notation.
The action of this operator on pλ when l(λ) > k is not specified by these conditions, but it
is determined.
If one wishes to give an expression for an operator that has the same action on pλ for
l(λ) ≤ k and the action on pλ for l(λ) > k is something else (say for instance 0), this is
possible by adding in terms of the form pµ pλ⊥ where l(λ) > k to CPa k .
86
3.
ZABROCKI
Homogeneous and elementary vertex operators
We require the following commutation relation between the homogeneous and monomial
symmetric functions.
Lemma 1 For k ≥ 0,
X
h⊥
m λ−(i) h ⊥
k mλ =
k−i
i≥0
m⊥
λ hk
=
X
h k−i m ⊥
λ−(i)
i≥0
where we will assume the convention m λ−(i) = 0 whenever λ − (i) is undefined.
Proof:
Note that hh i⊥ m λ , h µ i = hm λ , h i h µ i = δµ,λ−(i) . Therefore
h i⊥ (m λ ) = m λ−(i)
(5)
and h i⊥ (m λ ) = 0 if λ does not have a part of size i. The first identity follows from Eq. (1).
The second identity is a restatement of the first since
­ ⊥
® ­
® X­
®
m λ h k P, Q = P, h ⊥
(6)
P, m λ−(i) h ⊥
k mλ Q =
k−i Q
=
X­
i≥0
h k−i m ⊥
λ−(i) P,
Q
®
(7)
i≥0
2
P
Define CH 1k to be the operator CH 1k = λ : `(λ)≤k (−1)|λ| e1k |λ m ⊥
λ , and the operator CE 1k
P
⊥
|λ| k
k
to be CE1 = λ : `(λ)≤k (−1) h 1 |λ f λ .
The vertex operator property that we prove for the homogeneous and elementary symmetric functions is
Theorem 2
If l(λ) ≤ k, then CH 1k h λ = h 1k |λ and CE1k eλ = e1k |λ .
Proof: The proof is a matter of showing that for k > 1 operator CH 1k and h n (considered
as an operator that consists of multiplication by h n ) has the commutation relation CH 1k h n =
h n+1 CH 1k−1 and CH 11 (h n ) = h n+1 .
X
CH 1k h n =
(−1)|λ| e1k |λ m ⊥
(8)
λ hn
λ:`(λ)≤k
The sum here is over λ with the number of parts less than or equal to k. Apply Lemma 1
and rearrange the terms in the sum.
X
X
CH 1k h n =
(−1)|λ| e1k |λ
h n−i m ⊥
(9)
λ−(i)
λ:`(λ)≤k
i≥0
87
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
=
X X
(−1)|λ|−i (−1)i e1k |λ h n−i m ⊥
λ−(i)
(10)
X X
(−1)|λ|−i (−1)i h n−i ei+1 e1k−1 |(λ−(i)) m ⊥
λ−(i)
(11)
λ:`(λ)≤k i≥0
=
λ:`(λ)≤k i≥0
In the last equation, there is an assumption that e1k−1 |(λ−(i)) = 0 if λ − (i) is undefined.
As long as k > 1, making the substitution λ → λ + (i) yields the equation:
Ã
=
n
X
(−1)i h n−i ei+1
!Ã
X
!
(−1)
λ:`(λ)≤k−1
i=1
|λ|
e1k−1 |λ m ⊥
λ
This is equal to = h n+1 CH 1k−1 using the well known relation
n ≥ 0. If k = 1 then
CH 11 (h n ) =
n
X
(−1)i h n−i ei+1 = h n+1
(12)
Pn
r
r =0 (−1) er h n−r
= 0 for
(13)
i=1
Notice also that CH 1k acting on 1, yields h k1 since only one term is not 0.
The corresponding result for the CE1k operator follows by noting that CE1k = ωCH 1k ω.
2
The action of CH 1k on h λ when l(λ) > k is not known. The sum in the formula for CH 1k
is only over partitions λ such that l(λ) ≤ k and by adding terms of the same form but with
l(λ) > k it is possible to modify the formula so that the action on the h λ when l(λ) > k is
0, but the formula will not be as simple.
It would be interesting to know the action of these vertex operators on other bases besides
the one that it adds a row and column to. For instance, actions of ek , h k , and pk are known on
the Schur basis, but what is the action of an operator that adds a column to the homogeneous,
elementary, or power basis when it acts on the Schur basis?
Note the following two formulas that relate CH 1k and CE1k .
CH 1k =
X
λ:l(λ)≤k
CE1k =
X
λ:l(λ)≤k
(−1)|λ| CE1k (eλ )m ⊥
λ
(14)
(−1)|λ| CH 1k (h λ ) f λ⊥
(15)
This is the first instance when a pair of operators share a relation like this, and it will
occur with pairs of the other operators that appear in this article. These relations fall under
the category of ‘eerie coincidences’ (since they are very unexpected and can probably be
explained on some higher dimensional plane).
88
4.
ZABROCKI
Monomial and forgotten vertex operators
The vertex operators for the monomial and forgotten symmetric functions require a few
identities.
l(µ)!
then for k ≥ 0, ek =
Lemma 2 Let rµ = (−1)|µ|−l(µ) n 1 (µ)!n
2 (µ)!...
Proof:
P
µ`k r µ h µ
2
[3] example I.2.20, p. 33
Lemma 3 For µ a partition with |µ| > 0,
if µ − ( j) does not exist then rµ−( j) = 0.
P
j≥0 (−1)
j
rµ−( j) = 0 where it is assumed that
Pk
j
Proof:
j=0 (−1) ek− j h j = 0. Now expand ek− j in terms of the homogeneous basis
using the last lemma and equate coefficients of h µ on both sides of the equation.
2
Lemma 4 For k ≥ 0, ek⊥ m λ =
undefined.
P
⊥
µ r µ m λ−(µ) ek−|µ|
where m λ−(µ) = 0 if λ − (µ) is
P
⊥
. The expansion of the ei⊥ in terms of h ⊥
Proof: By Eq. (2) ek⊥ m λ = i≥0 ei⊥ (m λ )ek−i
µ is
given in the last lemma and so we have that
Ã
Ã
!
!
X X
X X
⊥
⊥
⊥
⊥
rµ h µ (m λ ) ek−i =
rµ m λ−(µ) ek−i
(16)
ek m λ =
i≥0
µ`i
i≥0
µ`i
by Eq. (5), and this is equivalent to the statement of the lemma.
2
P
Lemma 5 For a > 0, m (a) m λ = i≥0 (1 + n a+i (λ))m λ−(i)+(a+i) where it is assumed that
m λ−(i)+(a+i) = 0 if λ − (i) is undefined.
Proof: For partitions µ of |λ| + a, one has that the coefficient of m µ in m (a) m λ is equal
to h ⊥
µ (m (a) m λ ).
⊥
⊥
We note that for all n ≥ 0 that h ⊥
n m (a) = m (a) h n + h n−a . Apply this to the expression
for the coefficient of m µ
h⊥
µ (m (a) m λ ) =
l(µ)
X
h⊥
µ−(µ j )+(µ j −a) (m λ )
(17)
j=1
This implies that for the coefficient to be non-zero that µ must be equal to λ with a part
(say of size i) pulled away and a part of size a + i added in. The coefficient will be the
number of times that a + i appears in the partition µ (one more time than it appears in the
partition λ).
2
The first vertex operator that is presented here for the monomial symmetric functions
adds a row but it also multiplies by a coefficient, but this operator provides an easy method
for obtaining an operator that does not have this coefficient.
89
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
Proposition 1
For a > 0, let RM a(1) =
P
⊥
i
i≥0 (−1) m (a+i) ei
then
RM a(1) m λ = (1 + n a (λ))m λ+(a)
Proof:
RM a(1) m λ =
X
(−1)i m (a+i) ei⊥ m λ
(18)
i≥0
Apply Lemma 4 to get
=
X
(−1)i m (a+i)
X
rµ m λ−(µ)
(19)
µ`i
i≥0
The sum over i and µ may be combined to form one sum over all partitions µ.
=
X
µ
(−1)|µ|rµ m (a+|µ|) m λ−(µ)
(20)
Now multiplying by a monomial symmetric function with one part has an expansion
given in Lemma 5.
=
XX
(−1)|µ|rµ (1 + n a+|µ|+ j (λ − (µ)))m λ−(µ)−( j)+( j+a+|µ|)
µ
(21)
j≥0
The terms indexed by the same monomial symmetric function may be grouped together
by letting ν = µ + ( j).
=
XX
(−1)|ν|− j rν−( j) (1 + n a+|ν| (λ − ((ν) − ( j))))m λ−(ν)+(a+|ν|)
ν
=
X
ν
But
P
j≥0
X
(−1)|ν|− j rν−( j)
(1 + n a+|ν| (λ))m λ−(ν)+(a+|ν|)
|ν|− j
rν−( j)
j≥0 (−1)
(22)
(23)
j≥0
= 0 if |ν| > 0 by Lemma 3. There is one term left.
= (1 + n a (λ))m λ+(a)
(24)
2
An expression for an operator that adds a row without a coefficient can be written in
terms of this operator.
Theorem 3 For a > 0 define RM a =
P
(1)
k (RM a )k+1
(h ak )⊥
k≥0 (−1)
(k+1)!
then RM a m λ = m λ+(a) .
90
ZABROCKI
Proof:
steps.
Apply the previous proposition to this formula and reduce using the following
¡ (1) ¢i+1
X
¡ i ¢⊥
i RM a
RM a m λ =
ha mλ
(−1)
(i + 1)!
i≥0
X
(n a (λ) + 1) . . . (n a (λ) + i + 1)
(−1)i
m λ−(a i )+(a i+1 )
=
(i + 1)!
i≥0
¶
µ
nX
a (λ)
i n a (λ) + 1
=
(−1)
m λ+(a)
i +1
i=0
= m λ+(a)
(25)
(26)
(27)
(28)
The last equality is true because for n > 0,
Pn
i−1 n
(i)
i=1 (−1)
= 1.
2
Notice that the action of the RM a operators on the monomial basis implies that RM a RM b
= RM b RM a . This property is difficult to derive just from the definition of the operator.
This expression for the operator RM a is a little unsatisfying since the computation of
(RM a(1) )i can be simplified. The following operator shows how RM a can be reduced to
closer resemble CH 1k . To add more than one row at a time to a monomial symmetric
function, the formula resembles the vertex operator that adds a column to the homogeneous
basis.
Proposition 2 For a > 0 and k ≥ 0, we have that
¡
RM a(k)
RM a(1)
=
k!
¢k
=
X
(−1)|λ| m a k |λ eλ⊥
λ
with the understanding that m a k |λ = 0 if a k |λ is undefined. It follows that RM a(k) m λ =
)m λ+(a k ) .
( n a (λ)+k
k
Proof: By induction on k, we will show that RM a(1) RM a(k) = (k +1)RM a(k+1) . It follows that
(RM a(1) )k = k!RM a(k) . Since (RM a(1) )k m λ = (n a (λ) + 1)(n a (λ) + 2) · · · (n a (λ) + k)m λ+(a k ) ,
a (λ)+k)
then RM a(k) m λ = (na (λ)+1)(na (λ)+2)···(n
m λ+(a k ) .
k!
X
X
RM a RM a(k) =
(−1) j m (a+ j) e⊥j
(−1)|λ| m a k |λ eλ⊥
(29)
λ
j≥0
Commute the action of e⊥j and m a k |λ using Lemma 4.
=
X
(−1) j m (a+ j)
λ
j≥0
=
X
X
(−1)|λ|
rµ m a k |λ−(µ) e⊥j−|µ| eλ⊥
XXX
(−1)|λ|+ j rµ m (a+ j) m a k |λ−(µ) e⊥j−|µ| eλ⊥
j≥0
λ
µ
(30)
µ
(31)
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
91
The formula for multiplying a monomial symmetric function with one part is given in
Lemma 5.
=
XXXX
(−1)|λ|+ j rµ n a+ j+l (a k | λ − (µ) − (l) + (a + l + j))
µ
λ
j≥0
l≥0
× m a k |λ−(µ)−(l)+(a+l+ j) e⊥j−|µ| eλ⊥
(32)
The next step is to change the sum over µ so that it includes the part of size l, this is
equivalent to making the replacement µ → µ − (l).
=
XXXX
(−1)|λ|+ j rµ−(l) n a+ j+l (a k | λ − (µ) + (a + l + j))
µ
λ
j≥0
l≥0
× m a k |λ−(µ)+(a+l+ j) e⊥j+l−|µ| eλ⊥
(33)
Let i = j + l, then the sum over j can be converted to a sum over i.
=
XXXX
(−1)|λ|+i−l rµ−(l) n a+i (a k | λ − (µ) + (a + i))
λ
µ
l≥0 i≥l
⊥
eλ⊥
× m a k |λ−(µ)+(a+i) ei−|µ|
(34)
Interchange the sum over i and the sum over l. Since l ≥ 0 and i ≥ l then i ≥ 0 and
0 ≤ l ≤ i.
=
i
XXXX
(−1)l rµ−(l) (−1)|λ|+i n a+i (a k | λ − (µ) + (a + i))
λ
µ
i≥0 l=0
⊥
eλ⊥
× m a k |λ−(µ)+(a+i) ei−|µ|
(35)
⊥
Notice that since ei−|µ|
is zero for all i < |µ|, then all terms are zero unless i ≥ |µ|.
=
i
XX X X
(−1)l rµ−(l) (−1)|λ|+i n a+i (a k | λ − (µ) + (a + i))
λ
µ i≥|µ| l=0
⊥
eλ⊥
× m a k |λ−(µ)+(a+i) ei−|µ|
(36)
The sum over l is equal to 0 as long as |µ| > 0 using Lemma 3.
=
XX
(−1)|λ|+i n a+i (a k | λ + (a + i))m a k |λ+(a+i) ei⊥ eλ⊥
λ
(37)
i≥0
Let the sum over λ include the part of size i, then λ = λ + (i).
=
X
λ
(−1)|λ|
X
i≥0
n a+i (a k+1 | λ)m a k+1 |λ eλ⊥
(38)
92
ZABROCKI
The sum over i is now independent of λ since
= (k + 1)
X
λ
P
i≥0
n a+i (a k+1 | λ) will always be k + 1.
(−1)|λ| m a k+1 |λ eλ⊥ = (k + 1)RM a(k+1)
(39)
2
It follows that the formula for RM a(k) can be substituted into Theorem 3 and this provides
a more reduced form of the first formula given for RM a .
Corollary 1 For a > 0, RM a =
P
P
|λ|+k
m a k+1 |λ eλ⊥ (h ak )⊥ .
λ (−1)
k≥0
Since the forgotten basis is related to the monomial basis by an application of the involution ω, then the formulas for the symmetric function operator that adds a row to the
forgotten symmetric functions follows immediately.
Corollary 2 For a > 0, RF a =
P
k≥0
P
|λ|+k k+1
k ⊥
f a |λ h ⊥
λ (ea )
λ (−1)
has the property that
RF a f λ = f λ+(a)
There exists an operator T−X of the same form as the operators that exist already in
this paper that has the property that T−X P[X ] = 0, if P[X ] is a homogeneous symmetric
function of degree greater than 0 and T−X 1 = 1. This means that the operator applied to
an arbitrary symmetric function has the property that it picks out the constant term of the
symmetric function.
Proposition 3
T−X =
Define the operator
X
(−1)|λ| sλ0 sλ⊥
λ
Then for any dual bases {aµ }µ and {bµ }µ (that is, haµ , bλ i = δλµ ), this is equivalent to
T−X =
X
(−1)|λ| ω(aλ )bλ⊥
λ
This operator has the property that T−X sλ = 0 for |λ| > 0 and T−X 1 = 1.
Proof:
T−X sµ =
X
(−1)|λ| sλ sλ⊥0 sµ
(40)
λ
This is exactly the same expression as formula ([3], p. 90, (I.5.23.1)) with the x variables
substituted for the y. This expression is 0 unless sµ = 1.
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
93
It requires very little to show that this operator can be given an expression in terms of
any dual basis.
T−X =
X
(−1)|λ| ω(sλ )sλ⊥
λ
=
X
(−1)|λ|
λ
=
XX
X
hω(sλ ), ω(bµ )iω(aµ )sλ⊥
(41)
(42)
µ`|λ|
(−1)|µ| ω(aµ )hsλ , bµ isλ⊥
(43)
µ λ`|µ|
=
X
µ
(−1)|µ| ω(aµ )bµ⊥
(44)
2
Note that T−X is actually a special case of a plethystic operator T Z P[X ] = P[X + Z ].
Fix a basis of the symmetric functions, {aµ }µ , we may talk about the symmetric function
linear operator that sends aµ to the expression dµ (where {dµ }µ is any family of symmetric
function expressions). We can say that this operator lies in the linear span of the operators
sλ sµ⊥ and an expression can be given fairly easily.
Corollary 3 (The everything operator) Let {aµ }µ be a basis of the symmetric functions
and {bµ }µ be its dual basis. Then an operator that sends aµ to the expression dµ is given by
{d }
E {aµµ} =
X
µ
dµ T−X bµ⊥
{d }
In other words we have that E {aµµ} acts linearly, and on the basis aµ it has the action
{d }
E {aµµ} aµ = dµ .
Proof: Note that when bµ⊥ acts on a homogeneous polynomial, the result is a homogeneous
polynomial of degree |µ| less. Therefore if |µ| > |λ|, then bµ⊥ aλ = 0. If |µ| < |λ| then
T−X bµ⊥ aλ = 0 since T−X kills all non-constant terms. When |µ| = |λ|, we have that
bµ⊥ aµ = δλµ and therefore, T−X bµ⊥ aλ = δλµ . This also implies that
X
µ
dµ T−X bµ⊥ aλ =
X
µ
dµ δλµ = dλ
(45)
2
This operator looks too general to be of much use, but using known symmetric function
identities it is possible to reduce and derive expressions for other operators. For instance,
the symmetric function operator that adds a column to the monomial symmetric functions
is a special case of this.
94
ZABROCKI
Theorem 4 For a > 0, let
µ
¶
X
|λ| n a (λ) + k
CM a k =
(−1)
m λ+(a k ) eλ⊥ ,
k
λ
then CM a k m λ = m a k |λ with the convention that m a k |λ = 0 if a k |λ is undefined.
Proof:
{m
k
}
We will reduce an expression for E {m λa} |λ to one for CM a k .
{m
}
E {m λa} |λ =
k
X
m a k |λ
λ
X
(−1)|µ| m µ eµ⊥ h ⊥
λ
(46)
µ
Let rλµ be the coefficient of eµ in h λ (by an application of the involution ω it is also the
coefficient of h µ in eλ ). Then the expression is equivalent to
X
X
X
m a k |λ
(−1)|µ| m µ eµ⊥
rλγ eγ⊥
(47)
=
µ
λ
γ `|λ|
Rearranging the sums this may be rewritten as
XXX
=
(−1)|µ| m a k |λrλγ m µ eµ⊥ eγ⊥
µ
λ
(48)
γ
It is possible to group all the terms that skew by the same elementary symmetric function
by making the substitution µ → µ − (γ ) since the sum over µ and γ are over partitions.
XXX
=
(−1)|µ|−|γ | m a k |λrλγ m µ−(γ ) eµ⊥
(49)
µ
λ
γ
P
|γ |
⊥
|λ| ⊥
Note that m µ−(γ ) = h ⊥
γ (m µ ) and
γ (−1) rλγ h γ = (−1) eλ .
XX
(−1)|µ| m a k |λ (−1)|λ| eλ⊥ (m µ )eµ⊥
=
µ
(50)
λ
Notice that the first part of this expression is exactly the operator RM a(k) acting exclusively
on m µ . We may then apply Proposition 2 and note that the expression reduces to the sum
stated in the theorem.
2
The symmetric function operator that adds a column (or a group of columns) to the
forgotten symmetric functions can be found by conjugating the CM a k operator by the
involution ω to derive the following corollary.
For a > 0, let
µ
¶
X
|λ| n a (λ) + k
=
(−1)
f λ+(a k ) h ⊥
λ,
k
λ
Corollary 4
CF a k
then CF a k f λ = f a k |λ with the convention that f a k |λ = 0 if a k |λ is undefined.
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
95
Remark Lemma 5 is not true for a = 0, therefore the proofs of Proposition 1 and Proposition 2 do not hold for a = 0. Something interesting can be said of these operators
in this case.
the calculation carefully, it is possible to see that if we set
P By following
⊥
|λ|
=
(−1)
m
RM (k)
λ eλ , then
λ:`(λ)≤k
0
RM (k)
0 mλ =
µ
¶
k − `(λ)
mλ
k
(51)
With the convention that n 0 (λ) = −`(λ), Theorem 4 and Corollary 4 and their proof make
sense.
The operator that adds a sequence of rows to the monomial symmetric functions and the
operator that adds a sequence of columns are related by a pair of formulas similar to in the
case of formulas (14) and (15). Notice that Proposition 2 and Theorem 4 say that
X
(−1)|λ| RM a(k) (m λ )eλ⊥
(52)
CM a k =
λ
RM a(k) =
X
(−1)|λ| CM a k (m λ )eλ⊥
(53)
λ
This is ‘eerie coincidence’ number two. The relation between these two operators is very
similar to the relation between CH 1k and CE1k but not exactly the same. Once again this is
unexpected and unexplained.
5.
Schur vertex operators
A symmetric function operator that adds a row to the Schur functions is given in [3] (p. 95–96
I.5.29.d) that is of the same flavor as the other vertex operators presented here.
Theorem 5 (Bernstein) Let RSa =
addition, RSa RSb = −RSb−1 RSa+1 .
P
⊥
i
i≥0 (−1) h a+i ei ,
then RSa sλ = sλ+(a) if a ≥ λ1 . In
Proof: Repeated applications of this operator yields expressions of the Jacobi-Trudi sort.
Use the relation RSa h k = h k RSa − h k−1 RSa+1 (which follows from [3] example (I.5.29.b.5)
and (I.5.29.d)), RSa (1) = h a and follow the proof of [3] (I.3.(3.4”) p. 43) which does not
actually require that the indexing sequence be a partition. It follows then that
RSs1 RSs2 · · · RSsn (1) = det|h s j − j+i |1≤i, j≤n
(54)
2
Conjugating this operator by ω produces an operator that adds a column to a Schur
symmetric function. We will show in this section that a nice expression exists for a formula
for an operator that adds a column to a Schur function, but with the property that the result
is 0 if the partition is longer than the column being added.
96
ZABROCKI
It follows from the commutation relation of the RSa , that there is a combinatorial method
for calculating the action of RSa on a Schur function when m < λ1 . Let htk (µ) be the integer
i such that µck = (µ2 − 1, µ3 − 1, . . . , µi − 1, µ1 + i − k, µi+1 , . . . , µl(µ) ) is chosen to
be a partition. This amounts to removing the first k cells from the border of µ. If it is not
possible to find such an i such that µck is a partition then say that µck is undefined.
Corollary 5 Let ν = λ + (a + k) where k ≥ λ1 − a (ν is λ resting on a sufficiently long
first row).
RSa sλ = (−1)htk (ν)−1 sνck
where it is assumed that sνck = 0 if νck does not exist.
The proof of this corollary is not difficult, just a matter of showing that the commutation
relation of RSa RSb agrees with this definition of νck and that the vanishing condition
exists because RSa RSa+1 = 0. This definition and corollary are useful in showing that
an expression for (RSa )k can be reduced to a form that is very similar to the other vertex
operators presented here.
Lemma 6 For a ≥ 0,
X
(−1)|λ| sa k |λ sλ⊥0
(RSa )k =
λ
with the convention that sa k |λ is 0 if a k |λ is undefined.
Proof:
By induction on k. The statement agrees with Theorem 5 for k = 1.
RSa (RSa )k =
X
X
(−1)i h a+i ei⊥
(−1)|λ| sa k |λ sλ⊥0
(55)
λ
i≥0
ei⊥ can be commuted with the Schur function to produce
=
i
X
X
X
⊥
⊥
(−1)i h a+i
(−1)|λ|
e⊥j (sa k |λ )ei−
j sλ0
λ
i≥0
Interchange the order of all of the sums.
XXX
⊥
⊥
=
(−1)|λ|+i h a+i e⊥j (sa k |λ )ei−
j sλ0
λ
(56)
j=0
(57)
j≥0 i≥ j
Make the substitution that i → i + j, changing the sum so that it is over all i ≥ 0 and
expand the product ei⊥ sλ⊥0 . The notation that γ /λ0 ∈ Vi means that γ differs from λ0 by a
vertical i strip (λ0j ≤ γ j ≤ λ0j + 1 and |γ | = |λ| + i).
=
X
XXX
(−1)|λ|+i+ j h a+i+ j e⊥j (sa k |λ )
sγ⊥
λ
j≥0 i≥0
γ /λ0 ∈Vi
(58)
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
97
Make the substitution γ → γ 0 so that the sum is over all partitions γ that differ from λ
by a horizontal i strip and rearrange the sums.
=
XX X
λ
(−1)|λ|+i
i≥0 γ /λ∈Hi
X
(−1) j h a+i+ j e⊥j (sa k |λ )sγ⊥0
(59)
j≥0
Now it is only necessary to notice that the sum over j is actually an application of the
Schur vertex operator acting exclusively on the symmetric function sa k |λ . Switch the order
of the sums over the partitions and expression becomes
=
X
γ
(−1)|γ |
X X
i≥0 λ:γ /λ∈Hi
RSa+i (sa k |λ )sγ⊥0
(60)
There is a sign reversing involution on these terms so that only one term in the sum over
i and λ survives, namely, sa k+1 |γ . If i = γ1 then RSa+γ1 (sa k |(γ −(γ1 )) ) = sa k+1 |γ .
Take any partition λ in this sum such that γ /λ is a horizontal strip of length less than γ1 .
If RSa+i (sa k |λ ) = 0, then this term does not contribute to the sum. If RSa+i (sa k |λ ) = sa k |ν
then ν = λ + (i + n)cn , where n = γ1 − i. There is a combinatorial statement that can be
made about partitions that satisfy this condition, this is a lemma stated in [4] (Lemma 3.15,
p. 34).
Lemma 7 There exists an involution Iγn on partitions µ such that µ/γ is a horizontal
n strip, µcn exists and γ 6= µcn with the property that htn (Iγn (µ)) = htn (µ) ± 1 and
µcn = Iγn (µ)cn .
This is exactly the situation here. Set µ = λ + (i + n) then µ/γ is a horizontal strip of
size |µ| − |γ | = |λ| + i + n − |γ | = n. The result then is that all terms cancel except for
the terms such that γ = λ + (i + n)cn or i = γ1 and RSa+i (sa k |λ ) = sa k+1 |γ .
The sum therefore reduces to
=
X
γ
(−1)|γ | sa k+1 |γ sγ⊥0
(61)
2
With this expression for the Schur function vertex operator, it is possible to reduce the
expression for the ‘everything operator’ that adds a column to the Schur functions but is
zero when the length of the indexing partition is larger than the height of the column being
added.
P
⊥
|λ|
k
Theorem 6 For a, k ≥ 0, let CSa k =
λ (−1) (RSa ) (sλ )sλ0 . This operator has the
property that CSa k sλ = sa k |λ if l(λ) ≤ k and CSa k sλ = 0 for l(λ) > 0.
98
ZABROCKI
Proof: Take the expression for the everything operator that adds a columns of height k
using the convention that sa k |λ is zero whenever a k | λ is undefined.
{s
}
E {sλa} |λ =
k
X
sa k |λ
λ
X
(−1)|µ| sµ sµ⊥0 sλ⊥0
(62)
µ
The coefficients of the expansion of sµ sλ in terms of Schur functions are well studied
and there exists formulas and combinatorial interpretations for their calculation. P
The only
ν
s
=
properties that we require here is that the coefficients
in
the
the
expression
s
µ λ
ν cλµ sν
P ν
ν
ν0
⊥
have the property that cλµ = cλ0 µ0 and sλ sν = ν cµλ sµ .
=
X
sa k |λ
λ
X
X 0
ν
⊥
(−1)|µ| sµ
cλµ
0 sν 0
µ
(63)
ν
0
ν
ν
Next, we rearrange the sums and make the substitution cλµ
0 = cλ0 µ .
=
X
sa k |λ
λ
X
X
(−1)|ν|−|λ|
cλν 0 µ sµ sν⊥0
ν
(64)
µ
Therefore the sum over µ is just an application of sλ⊥0 on (sν ) and the sums can be
rearranged.
X
X
=
(−1)|ν|
(−1)|λ| sa k |λ sλ⊥0 (sν )sν⊥0
(65)
ν
λ
The sum over λ is now exactly an application of Lemma 6.
X
(−1)|ν| (RSa )k (sν )sν⊥0
=
(66)
ν
This is the expression given in the statement of the theorem.
2
The last of the ‘eerie coincidences’ of this article is that the CSa k and (RSa )k are related
by a pair of formulas similar to the case of formulas (14), (15) and (52), (53).
X
CSa k =
(−1)|λ| (RSa )k (sλ )sλ⊥0
(67)
λ
(RSa )k =
X
(−1)|λ| CSa k (sλ )sλ⊥0
(68)
λ
They say that once is happenstance, twice is coincidence and three times is a conspiracy.
This relationship can be made more explicit and it explains why these operators come in
pairs, but not why column adding operators happen to be related row adding operators for
both the Schur and monomial bases and why a similar relation exists with the homogeneous
and elementary vertex operators.
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
99
Let V be a linear operator from the space of symmetric functions to itself. Define
V̄ =
X
λ
(−1)|λ| V (sλ )sλ⊥0 =
X
(−1)|λ| V (aλ )(ωbλ )⊥
(69)
λ
where the sum is over all partitions λ and {aλ }λ and {bλ }λ are any two dual bases.
=
It is not difficult to show that V = V and that Eqs. (52), (53), (67), (68) may be summarized
as RM a(k) = CM a k and (RSa )k = CSa k . The relationship between (14) and (15) is not exactly
the same, but it follows that CH 1k (h λ ) = CE1k (h λ ) for all `(λ) ≤ k.
6.
An application: The tableaux of bounded height
One observation about the operator CSa k that could have an interesting application is that
CS0 k sλ = 0 if l(λ) > k and CS0 k sλ = sλ if l(λ) ≤ k. Knowing this and the commutation
calculate the number of pairs of standard tableaux
relation between RSa and h k allows us toP
of the same shape of bounded height [1] λ`n : l(λ)≤k f λ2 (where f λ is the number of standard
tableaux of shape λ).
Proposition 4 Let CP(n, k) be the collection of sequences of non-negative integers of
length k such that the sum is n.
X
λ`n : l(λ)≤k
f λ2
µ ¶Q
n
i< j (s j + j − (si + i))
n!
=
Qk
s
s∈CP(n,k)
i=1 (si + i − 1)!
X
The formula follows by applying
CS0 k to the symmetric function h n1 to arrive at a formula
P
for the symmetric function λ`n : l(λ)≤k f λ sλ .
Lemma 8
µ ¶
n
f λ sλ =
det |h s j − j+i |1≤i, j≤k
s
s∈CP(n,k)
X
X
¡ ¢
CS0 k h n1 =
λ`n:l(λ)≤k
Proof: Use the relation RSa h k = h k RSa − h k−1 RSa+1 , RSa 1 = h a and induction to calculate that
n
¡ ¢ X
RSk0 h n1 =
X
µ
(−1)n−l h l1
l=0 s∈CP(n−l,k)
¶
n
det |h s j − j+i |1≤i, j≤k
l, s
(70)
n ) f h n−|λ| we have that
Using the relation that sλ⊥ (h n1 ) = ( |λ|
λ 1
¡ ¢ X
¡ ¢
CS0 k h n1 =
(−1)|λ| (RS0 )k (sλ )sλ⊥0 h n1
λ
(71)
100
ZABROCKI
µ ¶
X
n
n−|λ|
(−1)|λ| (RS0 )k (sλ )
fλ h 1
|λ|
λ
µ ¶
n X
X
n
(−1)i
(RS0 )k ( f λ sλ )h n−i
=
1
i
i=0 λ`i
µ ¶
n
X
¡ ¢
n
(−1)i
(RS0 )k h i1 h n−i
=
1
i
i=0
=
(72)
(73)
(74)
Now using (70) we can reduce this further to
=
m
n X
X
X
µ ¶µ ¶
n
m
h n+l−m det |h s j − j+i |1≤i, j≤k
m l, s 1
(−1)l
m=0 l=0 s∈CP(m−l,k)
(75)
Now switch the sums indexed by l and m and then make the replacement m → m + l
=
n−l
n X
X
X
µ
(−1)l
l=0 m=0 s∈CP(m,k)
n
m +l
¶µ
¶
m + l n−m
h 1 det |h s j − j+i |1≤i, j≤k
l, s
(76)
Now switch the sums back and rearrange the binomial coefficients
=
n n−m
X
X
X
µ
(−1)l
m=0 l=0 s∈CP(m,k)
n
n − m, s
¶µ
¶
n − m n−m
h 1 det |h s j − j+i |1≤i, j≤k
l
(77)
Pn−m
(−1)l ( n−m
) will always be zero unless n − m = 0 and if n = m
Now the sum l=0
l
then it is 1 and so the entire sum collapses to
=
µ ¶
n
det |h s j − j+i |1≤i, j≤k
s
s∈CP(n,k)
X
(78)
2
Proof of Proposition 4: The proposition follows from this lemma with a little manipulation. There is a linear and multiplicative homomorphism that sends the symmetric functions
to the space of polynomials in one variable due to Gessel defined by θ (h n ) = x n /n!. This
homomorphism has the property that θ (sλ ) = f λ x |λ| /|λ|!. The image of the formula in the
lemma is then
Ã
!
X
X
¡
¡ n ¢¢
xn
θ CS0 k h 1 = θ
f λ sλ =
f λ2
(79)
n!
λ`n:l(λ)≤k
λ`n:l(λ)≤k
Therefore if we set (a)0 = 1 and (a)i = a(a −1) · · · (a −i +1) then we have (by making a
slight transformation that reverses order of the sequence first. . . j → n+1− j, i → n+1−i
VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS
101
and si → sn+1−i ) that
X
λ`n:l(λ)≤k
X
λ`n:l(λ)≤k
¯
¯
µ ¶
¯ (s j + j − 1)i−1 ¯
n
¯
¯
=
n!
det ¯
¯
(s
+
j
−
1)!
s
j
1≤i, j≤k
s∈CP(n,k)
X µn ¶ det |(s j + j − 1)i−1 |1≤i, j≤k
2
fλ =
n!
Qk
s
s∈CP(n,k)
i=1 (s j + j − 1)!
f λ2
X
(80)
(81)
The determinant is a specialization of the Vandermonde determinant in the variables
2
s j + j − 1 so the formula reduces to the expression stated in the proposition.
We note that in the case that k = 1 this sum reduces to 1 and in the case that k = 2 we
have that
X
λ`n : l(λ)≤2
f λ2 =
n µ ¶
X
n
j=0
n µ ¶2
X
n − 2j + 1
n n − 2j + 1
n! =
n− j +1
j ( j)!(n − j + 1)!
j
j=0
(82)
And this is an expression for the Catalan numbers. It would be interesting to see if these
expressions and equations could be q or q, t analogued.
References
1. F. Bergeron, L. Favreau, and D. Krob, “Conjectures on the enumeration of tableaux of bounded height,” Discrete
Mathematics 139 (1995), 463–468.
2. N. Jing, “Vertex operators and Hall-Littlewood symmetric functions,” Adv. Math. 87 (1991), 226–248.
3. I.G. Macdonald, Symmetric Functions and Hall Polynomials, Oxford Mathematical Monographs, Oxford UP,
2nd ed., 1995.
4. M.A. Zabrocki, “A Macdonald vertex operator and standard tableaux statistics for the two-column (q, t)-Kostka
coefficients,” Electron. J. Combinat. 5 (R45) (1998), 46.
5. A.V. Zelevinsky, Representations of Finite Classical Groups: A Hopf Algebra Approach, Springer Lecture
Notes, 869, 1981.