Vertex Operators for Standard Bases of the Symmetric Functions

Transcription

1 Journal of Algebraic Combinatorics 13 (2001, 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 duquébec 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 {p λ [X]} λ, Schur {s λ [X]} λ, monomial {m λ [X]} λ, homogeneous {h λ [X]} λ, elementary {e λ [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 λ i. The length of a partition λ is the largest i such that λ i 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 λ and is the partition such that λ i = the number of j such that λ j is greater than or equal to i. There is a standard inner product on symmetric functions p λ, p µ =z λ δ λµ where δ xy = 1 if x = y and 0 otherwise and z λ = i 1 i n i (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 (λ 1 + a,λ 2 +a,...,λ k +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,...i l(µ } {1,2,...l(λ} such that µ = (λ i1,λ i2,...,λ il(µ. If this sequence does not exist then λ (µ is again undefined.

2 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 (λ 1,λ 2,...,λ l(λ,µ 1,µ 2,..., µ l(µ into a partition. Occasionally within sums we will have an expression such as m λ (µ or h 1 k 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 {a λ } λ and {b λ } λ are dual if they have the property that a λ, b µ =δ λµ. By definition, the power symmetric functions are dual to the basis {p λ /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 ( s λ, s µ =δ λµ. There exists an involution, ω, on symmetric functions that relates the elementary and homogeneous bases by ωh µ = e µ and the monomial and forgotten bases by ωm µ = f µ.it also has the property that ωs λ = s λ. 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 f P, Q = P,fQ. Its action on an arbitrary symmetric function P may be calculated by the formula f P = λ P, fa λ b λ for any dual bases {a λ } λ and {b λ } λ. Using results and notation in [3] (p example (I.5.25, by setting f = µ (a µ f b µ where {a µ } µ and {b µ } µ are any dual bases. It follows that if f = i c i d i then f (PQ = i c i (Pd i (Q. Using this result and considering f to be multiplication by some symmetric function we have, h k f = i ( h i f h k i (1 e k f = i ( e i f e k i (2 p k f = ( p k f + fp k (3 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 g i } i where f i and g i 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.

3 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 85 The existence of such operators for the Schur ([3] p , [5] p. 69, and (row only Hall-Littlewood ([3] p , [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 e k e λ = e λ+(k, h k h λ = h λ+(k and p k p λ = p λ+(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 +kδ kj. This can be used to show the slightly more general relation pλ p k = p k pλ + kn k(λpλ (k (where it is assumed that pλ (k = 0ifλdoes not contain a part of size k. The vertex operator is given by the following theorem Theorem 1 For a 0 and k 0 define the following linear operator CP a k = λ : l(λ k p k l(λ a l(λ (p λi +a p λi p a pλ / zλ i=1 where the sum is over all partitions λ with less than or equal to k parts (if k = 0 then CP a 0 = 1.CP a k has the property that CP a k p µ = p a 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 CP a k 1 = pa k since p λ kills 1 for λ > 0. From the commutation relation of pλ and p k we derive that CP a k p j = (p j+a p j p a CP a k 1 + p j CP a k (4 The proof by induction follows from this relation. The formula CP a 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 p λ 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 p λ for l(λ k and the action on p λ for l(λ > k is something else (say for instance 0, this is possible by adding in terms of the form p µ p λ where l(λ > k to CP a k.

4 86 ZABROCKI 3. Homogeneous and elementary vertex operators We require the following commutation relation between the homogeneous and monomial symmetric functions. Lemma 1 For k 0, h k m λ = i 0 m λ (i h k i m λ h k = 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 h i m λ, h µ = m λ,h i h µ =δ µ,λ (i. Therefore h i (m λ = m λ (i (5 and hi (m λ = 0ifλdoes not have a part of size i. The first identity follows from Eq. (1. The second identity is a restatement of the first since m λ h k P, Q = P, h k m λq = P, mλ (i h k i Q (6 i 0 = i 0 hk i m λ (i P, Q (7 Define CH 1 k to be the operator CH 1 k = λ : l(λ k ( 1 λ e 1 k λm λ, and the operator CE 1 k to be CE 1 k = λ : l(λ k ( 1 λ h 1 k λ fλ. The vertex operator property that we prove for the homogeneous and elementary symmetric functions is Theorem 2 If l(λ k, then CH 1 k h λ = h 1 k λ and CE 1 k e λ = e 1 k λ. Proof: The proof is a matter of showing that for k > 1 operator CH 1 k and h n (considered as an operator that consists of multiplication by h n has the commutation relation CH 1 k h n = h n+1 CH 1 k 1 and CH 1 1(h n = h n+1. CH 1 k h n = ( 1 λ e 1 k λm λ h n (8 λ:l(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. CH 1 k h n = ( 1 λ e 1 k λ h n i m λ (i (9 λ:l(λ k i 0

5 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 87 = ( 1 λ i ( 1 i e 1 k λh n i m λ (i (10 λ:l(λ k i 0 = λ:l(λ k i 0 ( 1 λ i ( 1 i h n i e i+1 e 1 k 1 (λ (im λ (i (11 In the last equation, there is an assumption that e 1 k 1 (λ (i = 0ifλ (iis undefined. As long as k > 1, making the substitution λ λ + (i yields the equation: ( ( n = ( 1 i h n i e i+1 ( 1 λ e 1 k 1 λm λ i=1 λ:l(λ k 1 (12 This is equal to = h n+1 CH 1 k 1 n 0. If k = 1 then using the well known relation n r=0 ( 1r e r h n r = 0 for CH 1 1(h n = n ( 1 i h n i e i+1 = h n+1 (13 i=1 Notice also that CH 1 k acting on 1, yields h k 1 since only one term is not 0. The corresponding result for the CE 1 k operator follows by noting that CE 1 k = ωch 1 k ω. The action of CH 1 k on h λ when l(λ > k is not known. The sum in the formula for CH 1 k 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 e k, h k, and p k 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 1 k and CE 1 k. CH 1 k = ( 1 λ CE 1 k (e λ m λ (14 λ:l(λ k CE 1 k = ( 1 λ CH 1 k (h λ fλ (15 λ:l(λ k 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.

6 88 ZABROCKI 4. Monomial and forgotten vertex operators The vertex operators for the monomial and forgotten symmetric functions require a few identities. µ l(µ l(µ! Lemma 2 Let r µ = ( 1 n 1 (µ!n 2 (µ!... then for k 0, e k = µ k r µh µ Proof: [3] example I.2.20, p. 33 Lemma 3 For µ a partition with µ > 0, j 0 ( 1 j r µ ( j = 0 where it is assumed that if µ ( j does not exist then r µ ( j = 0. Proof: k j=0 ( 1 j e k j h j = 0. Now expand e k j in terms of the homogeneous basis using the last lemma and equate coefficients of h µ on both sides of the equation. Lemma 4 undefined. For k 0, e k m λ = µ r µm λ (µ e k µ where m λ (µ = 0 if λ (µ is Proof: By Eq. (2 ek m λ = i 0 e i (m λ ek i. The expansion of the e i in terms of h µ is given in the last lemma and so we have that ek m λ = ( r µ h µ (m λ ek i = ( r µ m λ (µ e k i (16 i 0 µ i i 0 µ i by Eq. (5, and this is equivalent to the statement of the lemma. Lemma 5 For a > 0, m (a m λ = i 0 (1 + n a+i(λm 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 (am 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 l(µ (am λ = 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. 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.

7 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 89 Proposition 1 For a > 0, let RM (1 a = i 0 ( 1i m (a+i e i then Proof: RM a (1 m λ = (1 + n a (λm λ+(a RM (1 a m λ = i 0 ( 1 i m (a+i e i m λ (18 Apply Lemma 4 to get = ( 1 i m (a+i r µ m λ (µ (19 i 0 µ i The sum over i and µ may be combined to form one sum over all partitions µ. = µ ( 1 µ r µ m (a+ µ m λ (µ (20 Now multiplying by a monomial symmetric function with one part has an expansion given in Lemma 5. = µ ( 1 µ r µ (1 + n a+ µ + j (λ (µm λ (µ ( j+( j+a+ µ (21 j 0 The terms indexed by the same monomial symmetric function may be grouped together by letting ν = µ + ( j. = ν = ν ( 1 ν j r ν ( j (1 + n a+ ν (λ ((ν ( jm λ (ν+(a+ ν (22 j 0 (1 + n a+ ν (λm λ (ν+(a+ ν ( 1 ν j r ν ( j (23 But j 0 ( 1 ν j r ν ( j = 0if ν >0 by Lemma 3. There is one term left. j 0 = (1 + n a (λm λ+(a (24 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 = k 0 (RM(1 a ( 1k k+1 (k+1! (h k a then RM a m λ = m λ+(a.

8 90 ZABROCKI Proof: steps. Apply the previous proposition to this formula and reduce using the following RM a m λ = i 0 i+1 ( RM (1 ( 1 i a (i + 1! ( h i a mλ (25 = ( 1 i (n a(λ (n a (λ + i + 1 m λ (a (i + 1! +(a i+1 i 0 (26 n a (λ ( = ( 1 i na (λ + 1 m λ+(a i + 1 i=0 (27 = m λ+(a (28 The last equality is true because for n > 0, n i=1 ( 1i 1 ( n i = 1. 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 (1 a i can be simplified. The following operator shows how RM a can be reduced to closer resemble CH 1 k. 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 (k a = ( RM (1 k a k! = λ ( 1 λ m a k λe λ with the understanding that m a k λ = 0 if a k λ is undefined. It follows that RM a (k m λ = ( n a(λ+k m k λ+(a k. Proof: By induction on k, we will show that RM a (1 RM(k a = (k+1rm a (k+1. It follows that (RM a (1k = k!rm a (k. Since (RM(1 a k m λ = (n a (λ + 1(n a (λ + 2 (n a (λ + km λ+(a k, then RM a (k m λ = (n a(λ+1(n a (λ+2 (n a (λ+k m k! λ+(a k. RM a RM a (k = j 0( 1 j m (a+ j e j ( 1 λ m a k λeλ (29 λ Commute the action of e j and m a k λ using Lemma 4. = ( 1 j m (a+ j ( 1 λ r µ m a k λ (µe j µ e λ (30 j 0 λ µ = ( 1 λ + j r µ m (a+ j m a k λ (µe j µ e λ (31 j 0 λ µ

9 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 91 The formula for multiplying a monomial symmetric function with one part is given in Lemma 5. = ( 1 λ + j r µ n a+ j+l (a k λ (µ (l + (a + l + j j 0 λ µ l 0 m a k λ (µ (l+(a+l+ je j µ e λ (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. = ( 1 λ + j r µ (l n a+ j+l (a k λ (µ + (a + l + j j 0 λ µ l 0 m a k λ (µ+(a+l+ je j+l µ e λ (33 Let i = j + l, then the sum over j can be converted to a sum over i. = ( 1 λ +i l r µ (l n a+i (a k λ (µ + (a + i λ µ l 0 i l m a k λ (µ+(a+iei µ e λ (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 ( 1 l r µ (l ( 1 λ +i n a+i (a k λ (µ + (a + i λ µ i 0 l=0 m a k λ (µ+(a+iei µ e λ (35 Notice that since ei µ is zero for all i < µ, then all terms are zero unless i µ. = i ( 1 l r µ (l ( 1 λ +i n a+i (a k λ (µ + (a + i λ µ i µ l=0 m a k λ (µ+(a+iei µ e λ (36 The sum over l is equal to 0 as long as µ > 0 using Lemma 3. = λ ( 1 λ +i n a+i (a k λ + (a + im a k λ+(a+iei eλ (37 i 0 Let the sum over λ include the part of size i, then λ = λ + (i. = λ ( 1 λ n a+i (a k+1 λm a k+1 λeλ (38 i 0

10 92 ZABROCKI The sum over i is now independent of λ since i 0 n a+i(a k+1 λ will always be k + 1. = (k + 1 λ ( 1 λ m a k+1 λe λ = (k + 1RM(k+1 a (39 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 = k 0 λ ( 1 λ +k m a k+1 λeλ (hk a. 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 RF a f λ = f λ+(a For a > 0, RF a = k 0 λ ( 1 λ +k f a k+1 λh λ (ek a has the property that 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 Define the operator T X = λ ( 1 λ s λ s λ Then for any dual bases {a µ } µ and {b µ } µ (that is, a µ, b λ =δ λµ, this is equivalent to T X = λ ( 1 λ ω(a λ b λ This operator has the property that T X s λ = 0 for λ > 0 and T X 1 = 1. Proof: T X s µ = λ ( 1 λ s λ s λ s µ (40 This is exactly the same expression as formula ([3], p. 90, (I with the x variables substituted for the y. This expression is 0 unless s µ = 1.

11 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 = λ = λ = µ = µ ( 1 λ ω(s λ sλ (41 ( 1 λ λ, ω(b µ ω(a µ sλ µ λ ω(s (42 ( 1 µ ω(a µ s λ,b µ sλ (43 λ µ ( 1 µ ω(a µ b µ (44 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, {a µ } µ, we may talk about the symmetric function linear operator that sends a µ to the expression d µ (where {d µ } µ is any family of symmetric function expressions. We can say that this operator lies in the linear span of the operators s λ sµ and an expression can be given fairly easily. Corollary 3 (The everything operator Let {a µ } µ be a basis of the symmetric functions and {b µ } µ be its dual basis. Then an operator that sends a µ to the expression d µ is given by E {d µ} {a µ } = µ d µ T X b µ In other words we have that E {d µ} {a µ } acts linearly, and on the basis a µ it has the action E {d µ} {a µ } a µ = d µ. 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 a λ = 0. If µ < λ then T X bµ a λ = 0 since T X kills all non-constant terms. When µ = λ, we have that bµ a µ = δ λµ and therefore, T X bµ a λ = δ λµ. This also implies that d µ T X bµ a λ = d µ δ λµ = d λ (45 µ µ 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.

12 94 ZABROCKI Theorem 4 CM a k = λ For a > 0, let ( 1 λ ( na (λ + k k m λ+(a k e 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: We will reduce an expression for E {m a k λ } {m λ } to one for CM a k. E {m a k λ } {m λ } = m a k λ ( 1 µ m µ eµ h λ (46 λ µ Let r λµ be the coefficient of e µ in h λ (by an application of the involution ω it is also the coefficient of h µ in e λ. Then the expression is equivalent to = m a k λ ( 1 µ m µ eµ r λγ eγ (47 λ µ γ λ Rearranging the sums this may be rewritten as = ( 1 µ m a k λr λγ m µ eµ e γ (48 λ µ 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. = ( 1 µ γ m a k λr λγ m µ (γ eµ (49 λ µ γ Note that m µ (γ = h γ (m µ and γ ( 1 γ r λγ h γ = ( 1 λ eλ. = ( 1 µ m a k λ( 1 λ eλ (m µeµ (50 µ 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. 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. Corollary 4 CF a k = λ For a > 0, let ( 1 λ ( na (λ + k k f λ+(a k h λ, 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.

13 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. By following the calculation carefully, it is possible to see that if we set RM (k 0 = λ:l(λ k ( 1 λ m λ eλ, then RM (k 0 m λ = ( k l(λ k m λ (51 With the convention that n 0 (λ = l(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 CM a k = λ ( 1 λ RM (k a (m λe λ (52 RM (k a = λ ( 1 λ CM a k (m λ e λ (53 This is eerie coincidence number two. The relation between these two operators is very similar to the relation between CH 1 k and CE 1 k 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 I.5.29.d that is of the same flavor as the other vertex operators presented here. Theorem 5 (Bernstein Let RS a = i 0 ( 1i h a+i ei, then RS a s λ = s λ+(a if a λ 1.In addition, RS a RS b = RS b 1 RS a+1. Proof: Repeated applications of this operator yields expressions of the Jacobi-Trudi sort. Use the relation RS a h k = h k RS a h k 1 RS a+1 (which follows from [3] example (I.5.29.b.5 and (I.5.29.d, RS a (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 RS s1 RS s2 RS sn (1 = det h s j j+i 1 i, j n (54 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.

14 96 ZABROCKI It follows from the commutation relation of the RS a, that there is a combinatorial method for calculating the action of RS a on a Schur function when m <λ 1. Let ht k (µ be the integer i such that µ k = (µ 2 1,µ 3 1,...,µ i 1,µ 1 +i k,µ i+1,...,µ l(µ is chosen to be a partition. This amounts to removing the first k cells from the border of µ. Ifitisnot possible to find such an i such that µ k is a partition then say that µ k is undefined. Corollary 5 first row. Let ν = λ + (a + k where k λ 1 a (ν is λ resting on a sufficiently long RS a s λ = ( 1 ht k(ν 1 s ν k where it is assumed that s ν k = 0 if ν k does not exist. The proof of this corollary is not difficult, just a matter of showing that the commutation relation of RS a RS b agrees with this definition of ν k and that the vanishing condition exists because RS a RS a+1 = 0. This definition and corollary are useful in showing that an expression for (RS a k can be reduced to a form that is very similar to the other vertex operators presented here. Lemma 6 For a 0, (RS a k = λ ( 1 λ s a k λs λ with the convention that s a k λ is 0 if a k λ is undefined. Proof: By induction on k. The statement agrees with Theorem 5 for k = 1. RS a (RS a k = ( 1 i h a+i ei ( 1 λ s a k λsλ (55 i 0 λ e i can be commuted with the Schur function to produce = ( 1 i h a+i ( 1 λ i 0 λ i e j (s a k λei j s λ (56 j=0 Interchange the order of all of the sums. = ( 1 λ +i h a+i e j (s a k λei j s λ (57 λ 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. The notation that γ/λ V i means that γ differs from λ by a vertical i strip (λ j γ j λ j + 1 and γ = λ +i. = ( 1 λ +i+ j h a+i+ j e j (s a k λ sγ (58 λ j 0 i 0 γ/λ V i

15 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 97 Make the substitution γ γ so that the sum is over all partitions γ that differ from λ by a horizontal i strip and rearrange the sums. = λ ( 1 λ +i i 0 γ/λ H i j 0 ( 1 j h a+i+ j e j (s a k λsγ (59 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 s a k λ. Switch the order of the sums over the partitions and expression becomes = γ ( 1 γ RS a+i (s a k λsγ (60 i 0 λ:γ/λ H i There is a sign reversing involution on these terms so that only one term in the sum over i and λ survives, namely, s a k+1 γ.ifi=γ 1 then RS a+γ1 (s a k (γ (γ 1 = s a k+1 γ. Take any partition λ in this sum such that γ/λis a horizontal strip of length less than γ 1. If RS a+i (s a k λ = 0, then this term does not contribute to the sum. If RS a+i (s a k λ = s a k ν then ν = λ + (i + n n, where n = γ 1 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 Iγ n on partitions µ such that µ/γ is a horizontal n strip, µ n exists and γ µ n with the property that ht n (Iγ n(µ = ht n(µ ± 1 and µ n = Iγ n(µ n. 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 n or i = γ 1 and RS a+i (s a k λ = s a k+1 γ. The sum therefore reduces to = γ ( 1 γ s a k+1 γ s γ (61 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. Theorem 6 For a, k 0, let CS a k = λ ( 1 λ (RS a k (s λ s λ. This operator has the property that CS a k s λ = s a k λ if l(λ k and CS a k s λ = 0 for l(λ > 0.

16 98 ZABROCKI Proof: Take the expression for the everything operator that adds a columns of height k using the convention that s a k λ is zero whenever a k λ is undefined. E {s a k λ } {s λ } = λ s a k λ ( 1 µ s µ sµ s λ (62 µ The coefficients of the expansion of s µ s λ in terms of Schur functions are well studied and there exists formulas and combinatorial interpretations for their calculation. The only properties that we require here is that the coefficients in the the expression s µ s λ = ν cν λµ s ν have the property that c ν λµ = cν λ µ and s λ s ν = ν cν µλ s µ. = λ s a k λ ( 1 µ s µ cλµ ν s ν (63 µ ν Next, we rearrange the sums and make the substitution c ν λµ = cν λ µ. = λ s a k λ ( 1 ν λ cλ ν µ s µsν (64 ν µ Therefore the sum over µ is just an application of sλ rearranged. on (s ν and the sums can be = ν ( 1 ν λ ( 1 λ s a k λs λ (s νs ν (65 The sum over λ is now exactly an application of Lemma 6. = ν ( 1 ν (RS a k (s ν s ν (66 This is the expression given in the statement of the theorem. The last of the eerie coincidences of this article is that the CS a k and (RS a k are related by a pair of formulas similar to the case of formulas (14, (15 and (52, (53. CS a k = λ ( 1 λ (RS a k (s λ s λ (67 (RS a k = λ ( 1 λ CS a k (s λ s λ (68 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.

17 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 99 Let V be a linear operator from the space of symmetric functions to itself. Define V = λ ( 1 λ V (s λ s λ = λ ( 1 λ V (a λ (ωb λ (69 where the sum is over all partitions λ and {a λ } λ and {b λ } 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 (RS a k = CS a k. The relationship between (14 and (15 is not exactly the same, but it follows that CH 1 k (h λ = CE 1 k (h λ for all l(λ k. 6. An application: The tableaux of bounded height One observation about the operator CS a k that could have an interesting application is that CS 0 k s λ = 0ifl(λ > k and CS 0 k s λ = s λ if l(λ k. Knowing this and the commutation relation between RS a and h k allows us to calculate the number of pairs of standard tableaux of the same shape of bounded height [1] λ n : l(λ k f λ 2 (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. λ n : l(λ k f 2 λ = s CP(n,k ( n i< j (s j + j (s i + i s k i=1 (s n! i + i 1! The formula follows by applying CS 0 k to the symmetric function h n 1 to arrive at a formula for the symmetric function λ n : l(λ k f λs λ. Lemma 8 ( CS 0 k h n 1 = λ n:l(λ k f λ s λ = s CP(n,k ( n det h s s j j+i 1 i, j k Proof: Use the relation RS a h k = h k RS a h k 1 RS a+1, RS a 1 = h a and induction to calculate that RS k ( n 0 h n 1 = l=0 s CP(n l,k ( n ( 1 n l h l 1 l, s det h s j j+i 1 i, j k (70 Using the relation that s λ (hn 1 = ( n λ f λh n λ 1 we have that ( CS 0 k h n 1 = ( 1 λ (RS 0 k (s λ sλ ( h n 1 λ (71

18 100 ZABROCKI ( n ( 1 λ (RS 0 k (s λ λ = λ n ( n = ( 1 i i = i=0 n λ i i=0 ( 1 i( n i Now using (70 we can reduce this further to = n m m=0 l=0 s CP(m l,k( 1 l( n m f λ h n λ 1 (72 (RS 0 k ( f λ s λ h n i 1 (73 (RS 0 k( h i 1 h n i 1 (74 ( m h n+l m 1 det h s l, s j j+i 1 i, j k (75 Now switch the sums indexed by l and m and then make the replacement m m + l = n n l l=0 m=0 s CP(m,k ( ( n m + l ( 1 l m + l l, s h n m 1 det h s j j+i 1 i, j k (76 Now switch the sums back and rearrange the binomial coefficients = n n m m=0 l=0 s CP(m,k ( ( ( 1 l n n m n m, s l h n m 1 det h s j j+i 1 i, j k (77 Now the sum n m l=0 ( 1l ( n m will always be zero unless n m = 0 and if n = m l then it is 1 and so the entire sum collapses to = ( n det h s s j j+i 1 i, j k (78 s CP(n,k 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(s λ = f λ x λ / λ!. The image of the formula in the lemma is then θ ( ( ( CS 0 k h n 1 = θ f λ s λ = fλ 2 x n (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+1then we have (by making a slight transformation that reverses order of the sequence first... j n+1 j,i n+1 i

19 VERTEX OPERATORS FOR SYMMETRIC FUNCTIONS 101 and s i s n+1 i that λ n:l(λ k λ n:l(λ k f 2 λ = f 2 λ = s CP(n,k s CP(n,k ( n det (s j + j 1 i 1 s (s j + j 1! n! 1 i, j k (80 ( n det (s j + j 1 i 1 1 i, j k s k i=1 (s n! j + j 1! (81 The determinant is a specialization of the Vandermonde determinant in the variables 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 = 2we have that λ n : l(λ 2 f 2 λ = n j=0 ( n n 2 j + 1 n ( n 2 j ( j!(n j + 1! n! = n 2 j + 1 j n j + 1 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, N. Jing, Vertex operators and Hall-Littlewood symmetric functions, Adv. Math. 87 (1991, I.G. Macdonald, Symmetric Functions and Hall Polynomials, Oxford Mathematical Monographs, Oxford UP, 2nd ed., 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, A.V. Zelevinsky, Representations of Finite Classical Groups: A Hopf Algebra Approach, Springer Lecture Notes, 869, 1981.