A BASIS FOR THE Y SUBSPACE OF DIAGONAL HARMONIC POLYNOMIALS JOSEPH ALFANO* Department of Mathematics, Assumption College 500 Salisbury Street, Worcester, Massachusetts 065-0005 ABSTRACT. The space DH n of S n diagonal harmonics is the collection of polynomials P (x; y) = P (x ;...; x n ; y ;...; y n ) which satisfy the dierential equations P n @r x i @ s y i P (x; y) = 0 for all r; s 0 (with r + s > 0). Computer explorations by Haiman have revealed that DH n has a number of remarkable combinatorial properties. In particular DH n is an S n module whose conjectured representation, graded by degree in y, is a sign twisted version of the action of S n on the parking function module. This conjecture predicts the character of each of the y-homogeneous subspaces Y j of DH n. The space Y 0 of diagonal harmonics with no y dependence is known in the classical theory. In this article we construct a basis for the subspace Y of diagonal harmonics linear in y. Using this basis we prove that the Y specialization of the Parking Function Conjecture is correct, and we provide a formula for the character of Y graded by degree in x. This last formula matches the Y specialization of a master conjecture of Garsia and Haiman which expresses the bigraded character of DH n.. Introduction Let Q[x; y] be the polynomial ring Q[x ; x ;... ; x n ; y ; y ;...; y n ] over the rationals. The diagonal action of the symmetric group S n on Q[x; y] is dened by setting, for = ( ; ;...; n ) S n and for P = P (x ; x ;... ; x n ; y ; y ;... ; y n ), P = P (x ; x ;...; x n ; y ; y ;... ; y n ) : : A polynomial P Q[x; y] is an S n diagonal invariant if P = P for all S n : Let I n be the ideal in Q[x; y] generated by the S n diagonal invariants, homogeneous of positive degree. Due to a theorem of Weyl [3], P a convenient set of generators for I n is given by the polarized power sum symmetric n functions xr i ys i (for r; s 0; with r + s n). We dene a bilinear form h; i on Q[x; y] by setting hp ; Qi = P (@ x ; @ x ;...; @ xn ; @ y ; @ y ;... ; @ yn ) Q(x ; x ;... ; x n ; y ; y ;... ; y n ) j x=y=0 : for monomials P; Q and extending by linearity; @ xi ; @ yi denote the dierential operators @=@x i ; @=@y i. This scalar product, known as the apolar form [3], is S n invariant, nondegenerate, and symmetric. It is easily shown that hx i P ; Qi = hp ; @ xi Qi and hy i P ; Qi = hp ; @ yi Qi : :3 * Work carried out under NSF grant support.
Thus multiplication by elements of Q[x; y] is adjoint to dierentiation. Dene DH n = I? n in Q[x; y] with respect to this scalar product. DH n is called the space of S n diagonal harmonic polynomials, since it equals the set fp Q[x; y] : @ r x i @ s y i P = 0 for r; s 0 (with r + s n)g : :4 In this paper we let Y j denote the subspace of elements of DH n which are homogeneous of degree j in y ; y ;... ; y n. In particular, Y 0 denotes the subspace of DH n whose elements depend only on x ; x ;... ; x n. P Now let E k denote the k th n polarization operator y i@ x k i, where k. Note that E k sends Y j into Y j+. To show this we need only verify that given any P DH n, E k P lies in DH n. Letting then Q be any diagonal invariant, we have Q ; E k P = Q ; y i @ k x i P = x k i @ y i Q ; P = 0 ; :5 the last equality due to the S n invariance of ( P n xk i @ y i )Q. Since hq; E k P i = 0 for all nontrivial homogeneous diagonal invariants Q, E k P must lie in the orthogonal complement of I n in Q[x; y], hence it must lie in DH n. Let (x) denote the Vandermonde determinant Q i<jn (x j?x i ): It is easy to check that (x) is a diagonal harmonic. Now it is well known that Y 0 is the linear span of the derivatives of (x). In Section we shall review some basic properties of Y 0 (see [], or [],[7], and [6]). In particular, we recall that Y 0 aords a graded version of the regular representation of S n. We then decompose Y 0 into submodules YS 0 Y 0 S and select a basis for each. These submodules respectively have character " Sn S and " Sn S, where " Sn S denotes the trivial character of the subgroup S induced up to S n, and denotes the sign character. In Section 3 we show that Y breaks into the direct sum of the images of Y 0 S under the polarization operators E ; E ;... ; E n?. This result completes our construction of a basis for Y and provides partial proof of the Operator Conjecture [8], which asserts that Y j is spanned by polarizations E k E k E kj of derivatives of (x). By our construction method in Section 3, it follows that Y has character (n? ) " Sn S. This value agrees with that predicted by the Parking Function Conjecture, which asserts that the S n character aorded by Y j equals (n?;n?;...;0) " Sn S m0 S m S mn? :6? where = (0 m0 m... (n? ) mn? n ) is a partition of P? j with m0 selected so that i m i = n. (See [8] for further details.) In Section 4 we give a formula for the character of Y graded by degree in x. We prove that this formula equals the Y specialization of a master conjecture of Garsia and Haiman [5] which expresses the bigraded character of DH n. The author is grateful to Mark Haiman, Adriano Garsia and Nolan Wallach for fruitful discussions, and assistance with some of the proofs presented herein.. Two subspaces of the S n harmonics Let a = (a ; a ;... ; a n ) be a point of Q n with distinct coordinates, and let [a] denote the orbit of a under the ordinary permutation action of S n. Let J [a] be the ideal in Q[x ; x ;... ; x n ] of
polynomials which vanish on [a]. The quotient Q[x ; x ;... ; x n ]=J [a] may be viewed as the coordinate ring of the orbit [a], hence the action of S n on this ring is equivalent to the regular representation. For any ideal J Q[x], we denote by gr J the ideal generated by the homogeneous components of highest degree of each of the elements of J. Observe that if J is invariant under the action of S n, so is gr J. It is easy to show that the actions of S n on the quotients Q[x]=J and Q[x]=(gr J) induce equivalent representations. Hence the action of S n on Q[x]=(gr J [a] ) is a graded version of the regular representation of S n. It develops that the quotients Q[x]=(gr J [a] ) are equivalent as graded S n modules for every a. This is a particular case of a general result concerning groups generated by reections []. In our case this is easy to show. Note that if P is a nontrivial homogeneous symmetric polynomial then P (x)? P (a) vanishes throughout [a], so P itself must belong to the ideal gr J [a]. In particular we have the containment (p ; p ;... ; p n ) gr J [a] : where p i denotes the i th power sum symmetric function. However, it is well known [] that the quotient Q[x ; x ;... ; x n ]=(p ; p ;... ; p n )? has dimension n! and this forces equality in.. Thus in? this case all the orthogonal complements gr J [a] with respect to the scalar product in. are the same. In fact, they coincide with the space Y 0 of solutions P of the dierential equations @ k x i P = 0 (k = ; ;...; n) : : For this reason the elements of Y 0 are usually referred to as the harmonics of S n. It is easy to see that the Vandermonde determinant (x) = detkx j? i k i;j=;:::;n :3 belongs to Y 0 and thus all of its derivatives do as well. A leading term argument easily shows that the n! derivatives @ x @ x @ x n n (x) (0 i i? ) :4 are independent, and thus they must give a basis for Y 0. In symbols we may write Y 0 = L [@ x (x)] :5 where L denotes the linear span. We take note that the equality in. may also be written as Y 0? = (p ; p ;... ; p n ) : :6 We are now ready to introduce the decomposition of Y 0 into submodules YS 0 Y 0 S. The subspace YS 0 will be obtained by a construction similar to that which gave us Y 0, and Y 0 S will be constructed after we discuss some of the basic properties of YS 0. Start here with the point b = (0; 0; ; ;...; n? ) and let J [b] and gr J [b] have the same denition as before. This given, we set Y 0 S =? gr J [b]? : :7 Theorem. YS 0 is an (n!=)-dimensional subspace of Y 0 which aords a graded version of the action of S n on the two-element subsets of f; ;...; ng. That is, the character of YS 0 is " Sn S. 3
Proof This is a special case of a known result concerning the harmonics of S n orbits (see Bergeron and Garsia [], or Garsia and Haiman [6]). For the sake of completeness we include the proof. Note rst that since all the power sum symmetric functions must belong to gr J [b] we necessarily have (p ; p ;...; p n ) gr J [b] : :8 Taking orthogonal complements gives that YS 0 is indeed a subspace of Y 0 as asserted. The common dimension of the spaces Q[x ; x ;...; x n ]=J [b] and Q[x ; x ;... ; x n ]=(gr J [b] ) and Y 0 S =? gr J [b]? is n!= since it must be equal to the cardinality of the orbit [b]. Note further that since the initial point b has S as a stabilizer, the action of S n on both quotients Q[x ; x ;... ; x n ]=J [b] and Q[x ; x ;... ; x n ]=(gr J [b] ) must be equivalent to the action of S n on the left S cosets. The same must hold true for the orthogonal complement of gr J [b]. Thus the last assertion follows from the fact that the S coset action is the same as the action on two-element subsets. The space Y 0 S can be represented in a manner similar to.5. For r = ; ;...; n let r (x) = detkx j? i k ;...;n;i6=r j=;...;n? : :9 In other words r (x) is the Vandermonde determinant in the variables x ;... ; x r? ; x r+ ;... ; x n. For any graded space V, we denote the Hilbert series by H V (q) = P i0 qi dim(v) i, which is the generating function of the dimensions of the subspaces (V) i homogeneous of degree i. Theorem. Our graded subspace Y 0 S is the linear span of derivatives of the polynomials r (x) for r = ;... ; n. The Hilbert series of Y 0 S is given by the polynomial H Y 0 S (q) = [n? ] q! n? where [k] q denotes the q-analog P k? i=0 qi and [k] q! = [] q [] q [k] q. i q i? :0 Proof Again as in the case of Theorem. a more general result can be found in [] or [6]. We give a sketch of the proof specialized to the present case. Consider the orbit [b] n? of b under the symmetric group S [;3;...;n] acting on the nal n? coordinates of b. It is obvious that the points of [b] n? also lie in [b]. Now let J [b]n? be the ideal of polynomials which vanish on [b] n?, and let gr J [b]n? denote the ideal generated by the homogeneous components of highest degree of elements of J [b]n?. We have the immediate ideal containment gr J [b]n? gr J [b]. Lastly set H [b]n? = (gr J [b]n? )?. We have now that H [b]n? YS 0. Observe that since the last n? coordinates of [b] are distinct, H [b]n? contains the harmonics of the symmetric group S [;3...;n]. In particular, the Vandermonde determinant in the last n? variables must belong to YS 0. But this is our determinant (x). Since YS 0 is S n invariant, all the S n images of (x) must also belong to YS 0. But these are equal (up to a sign change) to the other determinants r (x). Finally since YS 0 is closed under dierentiation, we have L [@ x r (x) : r = ; ;...; n] Y 0 S : : 4
To convert. into an equality we shall resort to a dimension argument. To be precise, since from Theorem. we know that dim YS 0 = n!=, we need only exhibit n!= independent derivatives of the determinants r (x). It develops that the following polynomials @ x @ x @ x r? r? @ x r+ r+ @ x n n r (x) 0 i i? for i r? : for r = ; 3;...; n ; 0 i i? 3 for r + i n are an instance in point. In fact, there are altogether r= n(n? ) (r? )!(r? ) (n? ) = (n? )! = n! of them and their leading monomials in the reverse lexicographic order are all distinct. This given, we can obtain the Hilbert series of YS 0 by calculating the generating function of the degrees of the polynomials in.. This computation yields the polynomial in.0 as asserted. :3 Now to construct Y 0 S we need some notation. For a given polynomial P Q[x] let ip P (x ; x ;...; x n ) = P (@ x ; @ x ;... ; @ xn ) (x) : :4 It is easy to show that ip is an automorphism of Y 0 as a vector space. This given, we set Theorem.3 The space Y 0 S The character of Y 0 S Y 0 S = ip Y 0 S : :5 is a graded S n module whose Hilbert series is given by the polynomial H Y 0 S (q) = [n? ] q! q n? n? is " Sn S. (n? i) q i? : :6 Proof From the denition.5 we derive that the Hilbert series of Y 0 S is related to that of Y 0 S by the identity H Y 0 S (q) = q (n ) HY 0 S (q? ) : :7 Carrying out this computation, using.0, yields.6 as desired. Let then f i (x)g ;;...;n!= be a homogeneous basis of YS 0 (the basis in. would do), and let A() = ka ij ()k ij be the matrix which expresses the action of a permutation on this basis. That is j (x) = n!= i (x) A ij () : :8 Since ip is a nonsingular map, the denition.5 gives that the collection f i (@ x )(x)g n!= must be a basis for Y 0 S. However, from.8 and the alternating property of the Vandermonde determinant, we derive that the action of on this basis is given by? n!=? j (@ x ) (x) = () i (@ x ) (x) A ij () : :9 5
Thus the character of the S n representation aorded by Y 0 S satises that char Y 0 S () = () n!= A ii () = () char Y 0 S () ; :0 and this completes our proof. We conclude this section with a proof of the basic fact that Y 0 = Y 0 S Y 0 S : : Since YS 0 and Y 0 S each have dimension n!=, it suces to show that these two subspaces are linearly independent. To achieve this end we shall need one more denition. Recall the operator ip, dened in.4, which is an automorphism of Y 0 as a vector space. This given, we dene the inverse map preip. For polynomials P and Q in Y 0 we set preip (Q) = P if and only if ip (P ) = Q : : It is clear that preip is also an automorphism of Y 0. Applying preip to our subspace Y 0 S, we observe the following inclusion. Lemma.4 preip Y 0 S Y 0 S? \ Y 0 : :3 Proof Recall our basis for YS 0 of @ x n? r (x), we see that given in.. Observing that the element r (x) is a constant multiple f@ x @ x @ r? x r? @ x n? r @ r+ x r+ @ n x n (x)g :4 is another basis for YS 0, where r and i satisfy the conditions given in.. It follows that a basis for preip YS 0 is given by elements of the form x x x r? r? x r n? x r+ r+ x n n + I :5 where I is an element of the kernel of ip in Q[x]. Since the kernel of ip P n is the ideal generated by the S n invariants xk i (k > 0), it is clear that I is orthogonal to YS 0. Now x x x r? r? x n? r x r+ r+ x n n is also orthogonal to every element in YS 0, due to the computation hx x x r? r? x r n? x r+ r+ x n n ; i = hx x x r? r? x r+ r+ x n n ; @ n? x r i = 0: :6 In fact @ x n? r is identically zero, due to Theorem. and the observation that the degree of x i in s (x) is no greater than n?. In view of Lemma.4, we have that preipy 0 S and Y 0 S are orthogonal subspaces of Y 0. It is immediate that these two subspaces are linearly independent. Now applying the ip operator to 6
these subspaces, and recalling that ip acts isomorphically on Y 0, we have that Y 0 S and Y 0 S are linearly independent. 3. Construction of a basis for Y Our goal in this section is to show that the space Y of diagonal harmonics homogeneous of degree one in y ; y ;...; y n breaks into the direct sum of n? copies of Y 0 S. More precisely we shall show that We start with the following. Y = Mn? k= E k Y 0 S : 3: Proposition 3. Each space E k Y 0 S is a graded S n module equivalent to Y 0 S, for k = ; ;...; n?. Proof We need to show that if f Y 0 S So let YS 0 and then we can have E k f = 0 only if f = 0 in the rst place. f = (@ x ) (x) : 3: Then E k f = Now the vanishing of E k f forces the equations y i @ k x i (@ x ) (x) : 3:3 @ k x i (@ x ) (x) = 0 (i = ; ;...; n) : 3:4 Dierentiating n?? k times with respect to x i yields that (@ x ) i (x) = 0 (i = ; ;...; n) : 3:5 However, using Theorem. we can derive from this that, as a dierential operator, must kill all the elements of YS 0. In particular it must kill itself. But that can only happen if it is identically zero. This shows that E k maps Y 0 S nonsingularly into Y. The fact that E k Y 0 S is equivalent to Y 0 S as a graded S n module is a simple consequence of the diagonal invariance and homogeneity of E k. To show the linear independence of the summands in 3. we need the following auxiliary fact. Lemma 3. If g = y i g i E k Y 0 S 3:6 then g i is of degree less than or equal to n?? k in the variable x i, and unless g vanishes, equality must hold for at least one i. 7
Proof From the expression given in 3.3 we get that, for some Y 0 S, g i = (@ x ) @ k x i (x) (i = ; ;...; n) : 3:7 Since (x) is of degree n? in each of the x i we immediately deduce our rst assertion. Assume now that the x i -degree of g i is strictly less than n??k for each i. This implies that the n??k th x i -derivative of g i must be identically zero. That is (@ x ) i (x) = 0 (i = ; ;...; n) : 3:8 In other words we are back in the same situation as in our previous proof. An element YS 0 kills all of the determinants i (x) and therefore by Theorem. must kill itself. This shows that and therefore also g must vanish. Proposition 3.3 E Y 0 S ; E Y 0 S ;...; E n? Y 0 S are independent submodules of Y. Proof Let g (k) E k Y 0 S for k = ; ;...; n? and suppose that Suppose that for some k n? we have g () + g () + + g (n?) = 0 : 3:9 g () = g () = = g (k?) = 0 : 3:0 If g (k) 6= 0 then by Lemma 3. g (k) would contain a monomial divisible by y i x n??k i for some i. Since such a monomial cannot be cancelled by any of the later summands g (k+) ;...; g (n?) (by Lemma 3. again), the whole sum in 3.9 could not vanish. Thus g (k) must vanish as well and by induction all the summands must separately vanish. Q.E.D Propositions 3. and 3.3 yield that the direct sum Mn? k= E k Y 0 S Y 3: has dimension precisely (n? )n!=. Thus to establish 3. and completely identify Y as an S n module we need only establish the following dimension inequality. The proof is due N. Wallach [], who stated it in a slightly dierent form. Theorem 3.4 dim Y (n? ) n! : 3: Before we can proceed with the proof we need to make some denitions and derive two auxiliary results. To begin we let W = y i h i : h i Y 0 ; h i = 0 : 3:3 8
It is clear that our space Y is a subspace of W. The following result identies it. Lemma 3.5 Y = y i h i : h i Y 0 ; @ x k i h i = 0 for k = 0; ;...; n? : 3:4 Proof Note that by denition an element h of the form h = is a diagonal harmonic if and only if it satises the dierential equations @ k x i h = 0 (for all k ) and y i h i 3:5 @ k x i @ yi h = 0 (for all k 0) : 3:6 This establishes 3.4. Now recall the operators ip, dened in.4, and preip, dened in.. The action of ip on W is dened by setting, for P n y ih i W, ip y i h i = y i ip h i : 3:7 It is easy to show that ip is an automorphism of W as a vector space. This given, we dene the action of preip on W. For polynomials P and Q in W, we set preip(q) = P if and only if ip (P ) = Q : 3:8 It is clear that preip is also an automorphism of W. Applying preip to our subspace of Y constructed in 3., we observe the following inclusion. Lemma 3.6 preip n? M k= E k Y 0 S Y? \ W : 3:9 Proof It suces to demonstrate that the elements preip E k (@ x ) (x) are orthogonal to Y for all YS 0 and all P k f; ;...; n? g. The element preip E k (@ x ) (x) may be expressed as n the projection of y ix k i (x) onto W in the following sense: preip E k (@ x ) (x) = preip y i @ k x i (@ x ) (x) = y i x k i (x) + I 3:0 where I is an element of the kernel of ip in Q[x; y]. Since the kernel of ip P n is the ideal generated by the S n invariants xk i (k > 0), it is clear that I is orthogonal to Y P n. Similarly since y i x k i (x) is a multiple of the diagonal invariant P n y i x k i, it too is orthogonal to Y. 9
In view of Lemma 3.6, we have that preip n? k= E ky 0 S and Y are orthogonal subspaces of W. Since preip acts isomorphically on W, we have that dim preip n? M k= E k Y 0 S = dim n? M k= E k Y 0 S = (n? ) n! : 3: Hence dim Y dim W? (n? )n!= : Finally we observe that dim W P n = (n?)n!. This holds true because the condition h i = 0 in the denition of W gives that dim W = (n? ) dim Y 0. From this consideration follows the inequality dim Y (n? )n!=, and Theorem 3.4 is proved. 4. The graded character of Y The equality in 3. together with Theorem.3 establishes that Y has character (n? ) " Sn S. This value agrees with that given by the Parking Function Conjecture [8], which predicts the character of each of the subspaces Y j. Our developments allow us now to construct an expression for the x-graded character of Y. Let V be a graded S n module and let (V) i denote the submodule of elements of V homogeneous of degree i. The graded character char q of V is dened by setting char q V = i0 q i char (V) i : 4: We let F denote the Frobenius (characteristic) map, which sends the irreducible representation to the Schur function s and extends homomorphically to the ring of symmetric functions [0]. It is easily seen that the Frobenius image of our ungraded character of Y is F char Y = (n? )! h (; n? ) = (n? ) e (; n? ) 4: where! denotes the fundamental involution on symmetric functions. We shall employ the -ring notation [5], a shorthand for q-analog symmetric function computations whereby: ) The alphabet x is written = x + x +..., and ) When dealing with a sum or product of alphabets, the power sum symmetric function p k [] is linear and multiplicative, i.e. p k [A + B] = p k [A] + p k [B] and p k [AB] = p k [A]p k [B]. For our purposes, we need only keep in mind the identities p k [(? q) ] = (? q k ) p k [] and p k [=(? q)] = (? q k ) p k[] 4:3 and extend homomorphically to the ring of symmetric functions Z Q(q). Theorem 4. The Frobenius image of the x-graded character of Y is given by the -ring identity F char q Y = [n? ] q! Q (; n? )[=(? q); q ] 4:4 where Q (x; q) is the Hall-Littlewood polynomial indexed by the partition. 0
Proof The graded character of Y 0 S is given by Bergeron and Garsia [] as char q Y 0 S = `n K;(; ~ )(q) = n? q (n? ) K ;(; n? )(q? ) ; 4:5 `n where ~K ; (q) = K ; (q? ) q n() ; 4:6 with K ; (q) denoting the Kostka-Foulkes polynomial and n() denoting the content of partition. Since ip complements degrees and conjugates the partition indexing the character, a simple computation based on 4.5 gives that char q Y 0 S = q n? `n 0 K ;(; n? )(q) : 4:7 Due to the decomposition 3., one sees immediately that char q Y = [n? ] q 0 K ;(; )(q) n? : 4:8 `n Equating Frobenius images of both sides of 4.8 gives F char q Y = [n? ] q s 0(x) K ;(; )(q) n? : 4:9 `n Since s 0 =! s we have (in -ring notation): F char q Y = [n? ] q! `n s [] K ;(; n? )(q) : 4:0 We recall from [0]* that Q (; n? )[; q] = `n s [(? q)] K ;(; n? )(q) : 4: Viewing this equation in terms of the power sum symmetric function basis fp [(? q) ] : ` ng, we apply the -ring identities from 4.3 to obtain Q (; n? )[=(? q); q] = `n s [] K ;(; n? )(q) : 4: Combining 4.0 and 4. completes our proof. Lastly we prove that our Theorem 4. conrms the Y specialization of a master conjecture of Garsia and Haiman. Their conjecture gives an expression for the Frobenius image of the bigraded character char q;t DH n = q i t j (DH n ) i;j 4:3 i;j0 * Although -ring notation is not used in [0], it is easy to translate Macdonald's equations (III.4.) and (III.4.7) to see that his S (x; q) equals our s [(? q)].
where (DH n ) i;j denotes the component of DH n homogeneous of degree i in variables x and degree j in y. To introduce the conjecture we shall need some notation. Let be a partition of n depicted in the French convention, i.e. justied along the west and south. Given any cell s in, we denote by l; l 0 ; a; and a 0 the parameters leg, coleg, arm, and coarm of the given cell, which are respectively the number of cells strictly north, south, east, and west of s in. This given, we set the following statistics of our partition. n() = s l 0 ~ h (q; t) = Y s (q a? t l+ ) n( 0 ) = s a 0 ~ h 0 (q; t) = Y s (t l? q a+ ) 4:4 B (q; t) = s q a0 t l0 (q; t) = Y s0;0 (? q a 0 t l0 ) where the product 0;0 is taken over all s in except the southwestern corner cell. Finally let ~ H (x; q; t) denote the Macdonald polynomials* (see [5], and also [4],[9]). We are ready to state the conjecture of Garsia and Haiman [5]. Conjecture 4. The Frobenius image of the x; y-bigraded character of DH n is given by F char q;t DH n = `n ~H (x; q; t) t n() q n(0 ) (? t) (? q) (q; t) B (q; t) ~h (q; t) ~ h 0 (q; t) : 4:5 To specialize this conjecture to Y, we examine the component of the right hand side of 4.5 linear in t. The summand corresponding to a given must be a multiple of t n(), since the denominator term ~ h (q; t) ~ h 0 (q; t) cannot be a multiple of t. Hence it suces to restrict our sum to partitions (n) and (n? ; ). We obtain ~H (n) (x; q; t) q (n ) Q n? (qi? t) + ~ H (n?;) (x; q; t) t q (n? ) (t + P n? i=0 qi ) (q n?? t ) (t? q n? ) Q n?3 (qi? t) 4:6 which equals ~H (n) (x; q; t) Q n? (? t q i )? P t n? ~H (n?;) (x; q; t) t ( q + n? (? t q ) (? n? q i ) t q n? ) Q n?3 (? t q i ) : 4:7 Applying the formal series expansion =(? u) = P j0 uj we see that the component of 4.7 linear in t equals that of n? ~H (n) (x; q; t) + t = ~ H (n) (x; q; t) + t n? q i q?i? ~ H(n?;) (x; q; t) t n? q i ~H (n) (x; q; t)? ~ H (n?;) (x; q; t) * Our notation ~ H (x; q; t) for the Macdonald polynomial follows that of [5]. Take note that the same object is denoted by H (; t; q) in [4]. : 4:8
Now it is known [4] that ~H (x; q; t)j t=0 = q n(0 ) `n s (x) K ; 0(q? ) : 4:9 In particular, we have the identities ~H (n) (x; q; t) = q (n ) `n s (x) K ;( n ) (q? ) ~H (n?;) (x; q; t)j t=0 = q (n? ) s (x) K ;(; n? )(q? ) : `n 4:0 The righthand sides of the equations in 4.0 equal F char q Y 0 and F char q Y 0 S by [], so it follows from. and.5 that ~H (n) (x; q; t)? ~ H (n?;) (x; q; t)j t=0 = F char q Y 0 S = q n? `n s 0(x) K ;(; n? )(q) : 4: We now have that the t term in 4.8 equals [n? ] q s 0(x) K ;(; )(q) n? ; 4: `n so this is the Frobenius image of the x-graded character of Y conjectured by 4.5. But this conjecture is immediately conrmed by equation 4.9. 5. References [] E. Artin, Galois Theory (University of Notre Dame, 944). [] N. Bergeron and A. Garsia, On certain spaces of harmonic polynomials, Contemporary Math. 38 (99) 5-86. [3] R. Ehrenborg and G.-C. Rota, Apolarity and canonical forms for homogeneous polynomials, Europ. J. Combinatorics 4 (993) 57-8. [4] A. Garsia and M. Haiman, A graded representation model for Macdonald's polynomials, Proc. Natl. Acad. Sci. USA 90 (993) 3607-360. [5] A. Garsia and M. Haiman, A remarkable q,t-catalan sequence and q-lagrange inversion, J. Alg. Combinatorics, to appear. [6] A. Garsia and M. Haiman, Orbit harmonics and graded representations, in: S. Brlek, ed., Publications du Lab. de Combinatoire et d'informatique Mathematique (Univ. du Quebec a Montreal), to appear. [7] A. Garsia and C. Procesi, On certain graded S n -modules and the q-kostka polynomials, Adv. Math. 94 (99) 8-38. [8] M. Haiman, Conjectures on the quotient ring by diagonal invariants, J. Alg. Combinatorics 3 (994) 7-76. [9] I.G. Macdonald, A new class of symmetric functions, Actes du 0 e Seminaire Lotharingien, Publ. I.R.M.A. Strasbourg 37/S-0 (988) 3-7. 3
[0] I.G. Macdonald, Symmetric Functions and Hall Polynomials, second edition (Clarendon, Oxford, 995). [] R. Steinberg, Dierential equations invariant under nite reection groups, Trans. Amer. Math. Soc. (964) 39-400. [] N. Wallach, personal communication, 99. [3] H. Weyl, The Classical Groups, Their Invariants and Representations, second edition (Princeton University, 946). 4