Combinatorics of Tesler matrices in the theory of parking functions and diagonal harmonics

Transcription

1 Journal of Combinatorics Volume 3, Number 3, , 202 Combinatorics of Tesler matrices in the theory of parking functions and diagonal harmonics D Armstrong, A Garsia, J Haglund, B Rhoades and B Sagan In [J Haglund, A polynomial expression for the Hilbert series of the quotient ring of diagonal coinvariants, Adv Math 227 (20) ], the study of the Hilbert series of diagonal coinvariants is linked to combinatorial objects called Tesler matrices In this paper we use operator identities from Macdonald polynomial theory to give new and short proofs of some of these results We also develop the combinatorial theory of Tesler matrices and parking functions, extending results of P Levande, and apply our results to prove various special cases of a positivity conjecture of Haglund Keywords and phrases: Parking functions, diagonal harmonics, tesler matrices Introduction Tesler matrices have been with us since Alfred Young s work at the beginning of last century, only it took Glenn Tesler s work on the higher order Macdonald operators to bring them to his attention and into the Sloane online encyclopedia of integer sequences To be precise, Tesler matrices emerge naturally from Young s Rising operator formula yielding the irreducible characters of the symmetric groups In fact, the Frobenius image of Young s formula (see [4, p 42]) may be written in the form () s μ [X] = n Ω[ i X] i<j n ( i / j ) μ n μ 2 n μn where, for any expression E, Ω[E] denotes the generating function of the complete homogeneous symmetric functions plethystically evaluated at E The first author is supported by NSF grant DMS The second author is supported by NSF grant DMS The third author is supported by NSF grant DMS The fourth author is partially supported by NSF grant DMS

2 452 D Armstrong et al (Readers unfamiliar with plethystic notation can consult [8, Chapter ]) More explicitly (2) Ω[E] = m 0 m h m [E] We should note also that Young s formula itself is but a specialiation of the Raising operator formula for Hall-Littlewood functions H μ (X; t) (see [4, p 23]) which may also be written in the form (3) H μ [X; t] = Since we may write (3) becomes (4) H μ [X; t] = n Ω[ i X] i<j n i / j t i / j μ n μ 2 i / j t i / j =Ω[ i j ( t)] n Ω[ i X] i<j n Ω[ i j ( t)] μ n μn n μ 2 n μn This in turn may be derived from N Jing s Vertex operator formula for Hall-Littlewood functions That is (5) H μ [X; t] =H μ H μ2 H μn where for any symmetric function F [X] weset (6) H a F [X] =F [X t ]Ω[X] a In fact, note that this gives H b H a =Ω[ (X t 2 )]Ω[ 2 X] =Ω[ a X]Ω[ 2 X]Ω[ 2 b 2 ( t)] and more generally we will have (7) H an H a2 H a = n Ω[ i X] i<j n which shows that (4) and (5) are identical Ω[ i j ( t)] a a b 2 a 2 2 n an

3 Combinatorics of Tesler matrices 453 Now note that the generic summand produced by the right-hand side of (7), after we expand all its factors according to (2) is n pi i h p i [X] This forces the equalities i<j n ( i / j ) pij h pij [ ( t)] a a 2 2 n an n s (8) p s + p sj p is = a s j=s+ (for all s n) We may thus associate to each summand an upper triangular matrix A = a i,j n i,j= by setting p i if i = j, a i,j = p ij if i<j, 0 otherwise Denoting by UP the collection of upper triangular matrices with non-negative integer entries, let us set for a given integral vector (a,a 2,,a n ) (9) T (a,a 2,,a n ) = { A = a i,j n i,j= UP: a s,s + n j=s+ s a s,j a i,s = a s s n } We will here and after refer to this as the collection of Tesler matrices with hook sums (a,a 2,,a n ) For a given matrix A = a i,j n i,j= UP set w HL (A) = n h ai,i [X] i<j n h ai,j [ ( t)] and call it the Hall-Littlewood weight ofa With this notation (5) becomes (0) H μ [X; t] = w HL (A) A T (μ n,μ n,,μ ) In a recent paper Haglund [9], through Macdonald polynomial manipulations, was led to a new expression for the Hilbert series of the space of

4 454 D Armstrong et al diagonal harmonics By using Sloane s On-Line Encyclopedia of Integer Sequences, he was able to identify this new expression as a weighted sum of Tesler matrices More precisely, Haglund obtained that () p n e n =( M )n w H (A) with (2) w H (A) = n h ai,i [ M] A T (,,,) i<j n where here and after it will be convenient to set (3) M =( t)( q) h ai,j [ M] In the same paper Haglund also proved that (4) p n m e n =( M )n w H (A) (for all m ) A T (,m,,m) It is now important to note that the same manipulations that yielded (0) but applied in reverse yield that, in view of (2), the Haglund identity in (4)canalsobewrittenintheform n (5) p n m e n =( M )n Ω[ i M] i<j n Ω[ i j M] m 2 m n Note further that as long as the entries (a,a 2,,a n ) are all positive each row of a matrix A T(a,a 2,,a n ) has to have at least one positive element, which is quite evident from the definition in (9) Thus, in view of (2), in spite of the denominator factor ( M) n, the expression (6) P a,a 2,,a n (q, t) =( M )n w H (A) A T (a,a 2,,a n) will necessarily evaluate to a polynomial Further experimentations revealed that when a a 2 a n the polynomial P a,a 2,,a n (q, t) turns out to have non-negative integer coefficients All these findings spurred a flurry of research activities aimed at the identification of the structures q, t counted by these polynomials and by what statistics In particular the formulas in

5 Combinatorics of Tesler matrices 455 () and (4) beg to be reconciled with the previously conjectured formulas [0, ] as weighted sums over parking functions and their natural m-extensions In this paper we present a variety of results which connect combinatorial properties of Tesler matrices to combinatorial properties of parking functions, diagonal harmonics and their closely related graded S n modules Our hope is that this work will help form a base for future research on connections between these topics We also give a new proof of the m =case of (5), as well as deriving some new results in the theory of plethystic Macdonald polynomial operators Our exposition will start with the basic symmetric function identities of the subject and then present the combinatorial developments that followed But before we can do that we need to collect a few identities of Macdonald polynomial theory which will turn out to be useful in our exposition Let us recall that in [5] we have set 2 Our manipulatorial tool kit (2) a) D k F [X] =F [ X + M ] Ω[ X] k b) Dk F [X] =F [ X M ] Ω[X] k for <k<+ Here k denotes the operation of taking the coefficient of k in the preceding expression, e m and h m denote the elementary and homogeneous symmetric functions indexed by m, and for convenience we have set (22) M =( t)( q), M =( /t)( /q) Recall that the symmetric function operator is defined by setting for the Macdonald basis { H μ (X; q, t)} μ H μ = T μ Hμ (with T μ = t n(μ) q n(mu ) and n(μ) = i (i )μ i) The operators in (2) are connected to and the polynomials H μ through the following basic identities: (23) (i) D 0 Hμ = D μ (q, t) H μ (i) D0 H μ = D μ (/q, /t) H μ (ii) D k e e D k = MD k+ (ii) Dk e e Dk = M D k+ (iii) e = D (iii) D = e (iv) = M D (iv) D = M

6 456 D Armstrong et al where e is simply the operator multiplication by e, denotes its Hall scalar product adjoint and (24) D μ (q, t) =MB μ (q, t) with B μ (q, t) = t l (c) q a (c) c μ Note that we also have (25) a) D k D k D k, b) Dk Dk D k Recall that for our version of the Macdonald polynomials the Macdonald Reciprocity formula states that (26) H α [ + ud β ] c α ( q utl a ) = Hβ [ + ud α ] c β ( q utl a ) (for all pairs α, β) We will use here several special evaluations of (26) To begin, canceling the common factor ( u) out of the denominators on both sides of (26) then setting u = gives (27) H α [MB β ] Π α = H β [MB α ] Π β (for all pairs α, β) On the other hand replacing u by /u and letting u = 0 in (26) gives (28) ( ) α H α [D β ] T α =( ) β H β [D α ] T β (for all pairs α, β) Since for β the empty partition we can take H β = and D β =, (26) in this case reduces to (29) Hα [ u ]= c α( n ut l q a )=( u) ( u) r e r [B μ ] This identity yields the coefficients of hook Schur functions in the expansion r=0 (20) Hμ [X; q, t] = λ μ s μ [X] K λμ (q, t)

7 Combinatorics of Tesler matrices 457 Recall that the addition formula for Schur functions gives { ( u) r ( u) if μ =(n r, r ), (2) s μ [ u] = 0 otherwise Thus (20), with X = u, combined with (29) gives for μ n (22) Hμ,s (n r,r ) = er [B μ ] and the identity e r h n r = s (n r,r ) + s (n r, r ) gives (23) Hμ,e r h n r = er [B μ ] Since for β =()wehave H β =andπ β =, formula (27) reduces to the surprisingly simple identity (24) Hα [M] =MB α Π α Last but not least we must also recall that we have the Pieri formulas (25) a) e Hν = d μν Hμ, b) e H μ = c μν Hν μ ν ν μ The final ingredients we need, to carry out our proofs, are the following summation formulas from [4] (the k = case of (27) occurs in unpublished work of M Zabrocki) (26) (27) c μν (q, t)(t μ /T ν ) k = ν μ d μν (q, t)(t μ /T ν ) k = μ ν { tq M h [ k+ Dμ (q, t)/tq ] if k, B μ (q, t) if k =0, { ( ) k e k [ Dν (q, t) ] if k, if k =0 Here ν μ simply means that the sum is over ν s obtained from μ by removing a corner cell and μ ν means that the sum is over μ s obtained from ν by adding a corner cell It will be useful to know that these two Pieri coefficients are related by the identity (28) d μν = Mc μν w ν

8 458 D Armstrong et al Recall that our Macdonald Polynomials satisfy the orthogonality condition (29) Hλ, H μ = χ(λ = μ)(q, t) The -scalar product, is simply related to the ordinary Hall scalar product by setting for all pairs of symmetric functions f,g (220) f,g = f,ωφg, where it has been customary to let φ be the operator defined by setting for any symmetric function f (22) φf[x] =f[mx] Note that the inverse of φ is usually written in the form (222) f [X] =f[x/m] In particular we also have for all symmetric functions f,g (223) f,g = f,ωg The orthogonality relations in (5) yield the Cauchy identity for our Macdonald polynomials in the form (224) Ω [ ɛ XY M ] = μ H μ [X] H μ [Y ] which restricted to its homogeneous component of degree n in X and Y reduces to (225) e n [ XY M ] = μ n H μ [X] H μ [Y ] Note that the orthogonality relations in (29) yield us the following Macdonald polynomial expansions Proposition 2 For all n we have [ a) e X ] n M = μ n H μ [X] [ b) h X ] [ X ] e k [B μ ] H μ [X] k w M en k M =, μ μ n

9 Combinatorics of Tesler matrices 459 c) h n [ X M ] = μ n T μ Hμ [X] d) ( ) n p n =( t n )( q n ) μ n Π μ Hμ [X] (226) e) e [X/M] n = μ n H μ [X] Hμ,e n f) e n = μ m H μ [X]MB μ Π μ Finally it is good to keep in mind, for future use, that we have for all partitions μ (227) T μ ω H μ [X;/q/, t] = H μ [X; q, t] Remark 2 It was conjectured in [4] and proved in [] that the bigraded Frobenius characteristic of the diagonal Harmonics of S n is given by the symmetric function (228) DH n [X; q, t] = μ n T μ Hμ (X; q, t)mb μ (q, t)π μ (q, t) (q, t) Surprisingly the intricate rational function on the right-hand side is none other than e n To see this we simply combine the relation in (24) with the degree n restricted Macdonald-Cauchy formula (225) obtaining (229) e n [X] =e n [ XM M ] = μ n H μ [X]MB μ Π μ This is perhaps the simplest way to prove (226)f) This discovery is precisely what led to the introduction of in the first place 3 The Hilbert series of diagonal harmonics Our first goal here is to obtain new proofs of the results in [9] by using the connection between Tesler matrices and plethystic operators The basic ingredient in this approach is provided by the following

10 460 D Armstrong et al Proposition 3 For any symmetric function F [X] and any sequence of integers a,a 2,,a n we have D an D a2 D a F [X] [ ] X=M n n = F M + Ω[ i M] Ω[ i j M] (3) = F [ M + i<j n n M i M i ] n ( i j )( tq i ( t i j )( q i i<j n ( i )( tq i ) ( t i )( q i ) j ) j ) a a 2 2 n an Proof The definition in (2)a) gives D a F [X] =F [X + M ]Ω[ X] and using (2)a) again we get D a2 D a F [X] =F [X + M + M 2 ]Ω[ (X + M 2 )]Ω[ 2 X] = F [X + M + M 2 ]Ω[ 2 M]Ω[ X 2 X] a a a 2 2 n an a a 2 2 a a 2 2 As before, this should make it clear that the successive actions of D a3 D ak will eventually yield the identity D an D a2 D a F [X] n M = F [X + i ] Ω[ i j M] i<j n Ω[ X 2 X n X] a a 2 2 n an Setting X = M gives the first equality in (6), but then the second equality holds as well since for any monomial v we have Ω[ vm] =Ω[vt + vq v qtv] = ( v)( qtv) ( tv)( qv) The identity in (3) has the following important corollary

11 Combinatorics of Tesler matrices 46 Theorem 3 (32) n p m e n =( M )n D n m D 0e n+ [ X M ] X=M Proof Setting a 0 =0,a 2 = a 3 = = a n = m andf [X] =e n+ [ X M ]in (3) gives (33) D n m D 0e n+ [ X M ] X=M = because in this case we have [ n M F M + i n Ω[ i M] i<j n Ω[ i ] [ = e n ] = 2 n j M] m 2 m n 2 n Combining (33) with (5) shows that (32) is simply another way of writing Haglund s identity () Our challenge here is to give a direct proof of (32) We have so far succeeded in carrying this out for m = Research to obtain the general case is still in progress The rest of this section is dedicated to the proof of the case m = of (32) and the derivation of another Tesler matrix formula for the Hilbert series of diagonal harmonics We will start with the latter formula since it requires the least amount of machinery Theorem 32 For all n we have (34) n p e n =( M )n D n p n [ X M =( M )n In particular it follows that n ] Ω[ i M] i<j n (35) n p e n =( M )n Proof The definition of D gives that Ω[ i j M] A T (n,,,, ) n 2 n w H [A] D p n [ X M ]=( p n [ X M ]+ n ) Ω[ X] =( ) n e n

12 462 D Armstrong et al Using this and (23)(iv) we derive that (36) ( M )n D p n n [ X M ]=( M )n D n e n = p n e n = p n e n This proves the first equality in (34) But Proposition 3 with a = a 2 = = a n = gives (37) D n p n [ X M ] X=M = ( + and (36) gives (38) n n i ) n Ω[ i M] n p e n =( M )n (+ n Ω[ i M] n n i ) i<j n i<j n Ω[ i j M] Ω[ i j M] 2 n 2 n Note next that if we set F = and a = a 2 = = a n = in (3) we obtain (39) 0 = D n n = Ω[ i M] Ω[ i X=M i<j n j M] 2 n that eliminates one of the terms in (38) We claim that only the term yielded by survives That is we have n (30) n Ω[ M i ] Ω[ M i / j ] = 0 (for all 2 k n) n k i<j n 2 n Let us now rewrite the LHS in the expanded form, that is n k n p i 0 pi i h p i [ M] i<j n r i,j 0 ri,j i ri,j j h ri,j [ M] 2 n The exponent of in the generic term of the product of these geometric series must satisfy the equation n p + r,j = j=2

13 Combinatorics of Tesler matrices 463 This is, of course impossible, causing (30) to be true precisely as asserted Using (39) and (30) in (38) proves the second equality in (34), completing our proof Our proof of (32) for m = is more elaborate and requires the following auxiliary identity Proposition 32 For any symmetric function F [X] we have MB μ Π μ (3) F [MB μ ]=Δ en F [X] X=M μ n+ Proof We need only prove this for F [X] = H γ [X] for arbitrary γ Inthis case (3) becomes (32) μ n+ MB μ Π μ Hγ [MB μ ]= H γ [M]e n [B γ ]=MB γ Π γ e n [B γ ] Since by the reciprocity in (27) we have (32) becomes Π γ μ n+ H γ [MB μ ] Π γ = H μ [MB γ ] Π μ, MB μ Hμ [MB γ ]=MB γ Π γ e n [B γ ], or more simply (33) μ n+ MB μ Hμ [MB γ ]=MB γ e n [B γ ] But recall that we have B μ = ν μ c μν and (3) becomes (34) μ n+ M Hμ [MB γ ] c μν = MB γ e n [B γ ] ν μ

14 464 D Armstrong et al Now for the left-hand side we have LHS = Mw ν c μν Hμ [MB γ ] w ν w ν n μ ν μ = d μν Hμ [MB γ ] w ν ν n μ ν = e [MB γ ] w H ν [MB γ ] ν ν n [ MBγ ] = e [MB γ ]e n = MB γ e n [B γ ]=RHS (!!!!) M and our proof is complete As a corollary we obtain Proposition 33 For F Λ =k with k n we have (35) μ n+ MB μ Π μ F [MB μ ]= { F [X] if k = n, X=M 0 if k<n Proof From (3) we get that the left hand side of (35) is Δ en F [X] X=M but for a symmetric function F [X] ofdegreen we have Δ en F [X] = F [X], thus the first alternative in (5) is immediate On the other hand for k<n the expansion of F in the Macdonald basis will involve H γ s with γ k and even before we make the evaluation at X = M the identity Δ en Hγ [X] =e n [B γ ] H γ [S] =0 (forallγ k<n) forces Δ en F = 0, yielding the second alternative in (35) We are now in a position to give a new and direct proof of (32) for m = Theorem 33 (36) n p e n = ( M ) nd n 0 e n+ [ X M ] X=M

15 Combinatorics of Tesler matrices 465 Proof From (226)a) and (23)(i) we get D n 0 e n+ [ X M ] X=M = ( D n 0 = = μ n+ n k=0 μ n+ H μ [X; q, t] ) X=M H μ [X; q, t]( MB μ ) n ( ) n ( M) k k μ n+ X=M MB μ Π μ B k μ, and Proposition 33 with F = e [ X M ] k and 0 k n gives ( M ) nd n 0 e n+ [ X M ] X=M = en [ X M ] X=M (by (226)e) and the definition of ) = μ n (by (24)) = μ n (by (226)f) and p = e )= n p e n T μ Hμ [X] Hμ,e n ] X=M T μ MB μ Π μ Hμ,e n This proves (36) and completes our argument Remark 3 Note that the identity in (32) for m = 2 becomes (37) n p 2 e n =( M )n D n D 0 e n+ [ X M ] X=M Since (23)(iii) gives D = e we can rewrite this as (38) n p 2 e n = ( M )n e n D 0 e n+ [ X M ] X=M which is a really surprising formula, since it shows that there are identities for these operators that still remain to be discovered Remark 32 Computer experimentation has shown that ( ) n p n m e n is a polynomial in Q[/q, /t] with non negative integer coefficients This phenomenon can be explained by means of the operator which is defined by setting for any symmetric polynomial F [X; q, t] F [X; q, t] =ωf[x;/q, /t]

16 466 D Armstrong et al In fact, note that for F = H μ [X; q, t] (227) gives (39) H μ [X; q, t] = ω H μ [X;/q, /t] = T μ Hμ [X; q, t] Thus (227) again gives H μ [X; q, t] = H μ [X; q, t] = T μ Hμ [X; q, t] = Hμ [X; q, t] and since the operator is linear in the linear span of the basis { H μ } μ it follows that (320) = Now this implies that ( ) n m e n [X] = m ( ) n h n [X] and the Schur positivity of ( ) n m e n [X] now follows from the conjectured Schur positivity of m ( ) n h n [X] Since the operators p and commute we see then that applying to both sides of (4) we derive that (32) p n m h n =( tq M )n w H (A) (for all m ) t=/t,q=/q A T (,m,,m) Remark 33 Note that using the operator we can derive from (226)f) that (322) h n = μ m ω H μ [X;/q, /t]( /t)( /q)b μ (/q, /t)π μ (/q, /t) (/q, /t) Since by definition we have (q, t) = c μ(q a(c) t l(c)+ )(t l(c) q a(c)+ ) Π μ (q, t) = c μc (0,0) ( q a (c) t l (c) )

17 Combinatorics of Tesler matrices 467 Thus (/q, /t) = c μ(q a(c) t l(c) )(t l(c) q a(c) ), Π μ (/q, /t) = c μc (0,0) ( q a (c) t l (c) ) (/q, /t) =(qt) n Tμ 2 (q, t), Π μ (/q, /t) =( ) n Tμ Π μ (q, t) and (322) becomes (323) h n =( qt) n μ m H μ [X; q, t]mb μ (/q, /t)π μ (q, t) (q, t) It should be noted that other surprising identities may be obtained by means of the operator For instance applying it to both sides of p Hμ (X; q, t) = ν μ c μν(q, t) H ν [X; q, t) gives by (39) p Hμ [X; q, t] = c μν (/q, /t) Hν [X; q, t] T μ T ν μ ν from which we derive that c μν (/q, /t) = Tν T μ c μν (q, t) An application of to the identity B μ (q, t) = ν μ c μν(q, t) gives (324) T μ B μ (/q, /t) = ν μ c μν (q, t)t ν which for μ n can also be rewritten as (325) e n [B μ (q, t)] = ν μ c μν (q, t)t ν 4 Further remarkable D 0 identities In this section we will explore some of the consequences obtained by combining the action of D o with the following identity of Haglund [7] Proposition 4 For any positive integers m, n and any P Λ n, we have (4) Δem e n,p = Δ ωp e m,s m

18 468 D Armstrong et al Proof The operator Δ em applied to both sides of (226) gives (42) Δ em e n = μ n MB μ Π μ Hμ (x; q, t) e m [B μ ] Using again (226) with n m we also have (43) Δ ωp e m = α m MB α Π α Hα (x; q, t) w α (ωp)[b α ] Using (42) and (43) we get the explicit form of (4), which is (44) μ n MB μ Π μ e m [B μ ] Hμ,P = MB α Π α (ωp)[b α ] w α α m This given, our idea here as in [7], is to establish (4) by checking its validity when P varies among all the members of a symmetric function basis It turns out that a simpler proof of (4) is obtained by testing (43) with the modified Macdonald basis { } ω Hγ [MX; q, t] w γ The source of the simplification is due to the fact that this basis is precisely the Hall scalar product dual of the Macdonald basis { Hγ [X]} γ Usingthis fact, putting P = ω H γ[mx] w γ in (44) gives γ MB γ Π γ e m [B γ ] w γ = α m MB α Π α w α Hγ [MB α ] w γ Carrying out the simplifications this may be rewritten as (45) B γ e m [B γ ]= α m and Macdonald reciprocity reduces this to (46) B γ e m [B γ ]= α m B α Π α w α Hγ [MB α ] Π γ B α w α Hα [MB γ ]

19 Combinatorics of Tesler matrices 469 Now we may rewrite (26) with k =0,as B α = β α c αβ Using this in the right hand side of (45) gives (47) RHS = H α [MB γ ] c αβ = w α α m β α Next we use (28), in the form β m w β α β H α [MB γ ]c αβ w β w α and (47) becomes RHS = M c αβ w β w α == M d αβ = M β m β m = B γ β m w β (by (226)a)) = B γ e m [ MBγ M α β H α [MB γ ]d αβ MB γ Hβ [MB γ ] w β H β [MB γ ; q, t] w β ] = B γ e m [B γ ] proving (45) as desired Study of the above proof led us to the following variation of (4) Proposition 42 (48) ( qt) m Δ hm e n,p = Δ ωp s m,s m Proof As we did for (4), we will prove (48) by testing it with P = ω H γ [MX]/w γ To this end note first that the left-hand side is LHS =( qt) [ ] m MB μ Π μ h m Bμ Hμ,P μ n and the right-hand side (using (323)) is RHS =( qt) m μ m M B μ Π μ (ωp)[b μ ]

20 470 D Armstrong et al With this choice of P we get LHS =( qt) m MB γπ γ h m [ Bγ ] w γ and RHS =( qt) m μ m M B μ Π μ T μ Hγ [MB μ ] w γ We are thus reduced to showing, upon omission of the common factor ( qt) m, division by MΠ γ /w γ and setting B μ = B μ (/q, /t), (49) B γ h m [ Bγ ] = μ m B μ Π μ T μ Hγ [MB μ ] Π γ The next idea is to use the identity in (324), namely T μ Bμ = ν μ c μν T ν and transform (49) to B γ h m [ Bγ ] = (by reciprocity) = μ m = ν m = ν m ν m = h [MB γ ] M Π μ Hγ [MB μ ] c μν T ν Π γ ν μ T ν w ν μ ν Π μ Hγ [MB μ ] w ν c μν Π γ T ν Π μ Hγ [MB μ ] d μν /M w ν Π μ ν γ T ν w ν μ ν ν m H μ [MB γ ]d μν /M which is plainly true, and completes the proof of (48) T [ ν MBγ ] Hν [MB γ ]=B γ h m w ν M An application of Proposition 4 yields another remarkable consequence of the action of the operator D 0

21 Combinatorics of Tesler matrices 47 Theorem 4 Let k 0 Recalling the notation in (6) for the Tesler matrix polynomials we have en+k,h k h n = P +a,+a 2,,+a n (q, t) a +a 2 + +an k 0 a i Proof An application of (4) with m = n + k + and F = h k h n gives Δen+ke n+k,h k h n = Δeke ne n+k+,h n+k+ Now (226)f) gives Δeke n e n+k+,h n+k+ = μ n+k+ (by the second case of (35)) = ( M )n MB μ Π μ e k [B μ ]Bμ n μ n+k+ MB μ Π μ e k [B μ ]( MB μ ) n =( M )n μ n+k+ H μ [X] e k [B μ ]( MB μ ) n X=M (by (23)) = ( M )n D n 0 μ n+k+ H μ [X] Hμ,e k h n+ X=M =( M )n D n 0 e n+ [ X M ] hk [ X M ] X=M On the other hand an application of Proposition 3 for a = a 2 = = a n = 0 and F = e n+ [ X M ] hk [ X M ] gives D0 n e n+h k X=M = e n+ [ + n ] [ n hk i + [ n ] n = h k + i Ω[ i M] = a +a 2 + +an k 0 a i n Ω[ i M] ] n i Ω[ i M] i<j n i<j n Ω[ i Ω[ i j M] j M] i<j n 2 n +a Ω[ i j M] +a n +an n

22 472 D Armstrong et al 5 From parking functions to Tesler matrices In a recent work P Levande [2] establishes a beautiful weight preserving bijection between parking functions and Tesler matrices with hook sums (,,,) and only one non ero element in each row We will soon be deriving more general forms of Levande s results, and as a means to that end we include our own exposition of part of his work here, in notation compatible with previous sections of this paper To state and prove his results we need to introduce some notation and establish a few auxiliary facts In our previous writings, parking functions were depicted as tableaux in the n n lattice square as in the figure below In this representation a parking function is constructed by first choosing a Dyck path, then placing circles in the cells adjacent to its NORTH steps and then filling the circles with, 2,,n in a column increasing manner The label u i, on the left of row i, gives the number of cells between the NORTH step in that row and the diagonal The label v i on the right of row i, gives the car in that row In the computer a PF is stored as the two line array [ ] v v PF = 2 v n u u 2 u n Thus, in the example above we obtain PF = [ ] It is easy to see that a necessary and sufficient condition for such an array to represent a PF is that we have u =0, u i u i + withu i = u i += v i >v i Using this representation the two statistics area and dinv can be simply defined by setting

23 Combinatorics of Tesler matrices 473 area(pf)= dinv(pf)= n u i, i<j n ( χ(ui = u j & v i <v j )+χ(u i = u j +&v i >v j ) ) However historically this construction of a PF was only devised so as to assure that the corresponding preference function did park the cars In fact in the original definition a parking function for a one way street with n parking spaces is a vector f =(f,f 2,,f n ) [,n] n such that #{j : f j i} i (for all i n) The simple bijection that converts the classical definition to its tableau form is simply to stack the cars as columns above the parking spaces they prefer to park Thus in the example above car wants to park in space 2, car 2 wants to park in space 6, etc So the correponding preference function turns out to be pf =(2, 6, 2,, 7,, 6, ) Of all the permutations we previously obtained from a parking function there are two more that we overlooked Namely the one which gives the history or better, the time order in which the places are filled and the other gives the arrangement of the cars after they are finally all parked It is easy to see that these permutations are inverses of each other, We will refer to the former as the filling permutation and the latter as the parking permutation For instance for the above parking function we have filling(pf) =(2, 6, 3,, 7, 4, 8, 5), parking(pf) =(4,, 3, 6, 8, 2, 5, 7) There is a simple algorithm for constructing filling(pf) from the preference function pf This is none other than an implementation of the way the parking takes place We need only illustrate it in our particular example Car arrives and parks in 2, car 2 arrives and parks in 6, but when car 3 arrives 2 is occupied and parks in 3 Car 4 arrives and parks in Car 5 arrives and parks in 7 But when car 6 arrives, who wants to park in, finds, 2, 3 occupied, thus ends up parking in 4 since it is still free The algorithm to get α = filling(pf) from pf, starts by setting α = pf[] then inductively having constructed the first i components of α and given that pf[i] =a

24 474 D Armstrong et al we let α i = a if a is not in (α,α 2,,α i )andletα i = a + k + if a, a +,,a+ k are in (α,α 2,,α i )anda + k +isnot Finally, the permutation parking(pf) is the inverse of filling(pf) since filling(pf)[i] gives where car i parks or equivalently where we must place i in parking(pf) A classical beautiful fact in the theory of parking functions, which goes back to Kreweras is Theorem 5 The inverse algorithm yields that for any α S n the polynomial (5) N α (q) = q area(pf) factors into a product of q-integers filling(pf)=α Proof Let α =(3, 6,, 4, 7, 2, 5) be the given filling permutation The tree depicted on the right gives all the preference functions that have α as their filling permutation To see this, note that when car arrives and parks at 3 it is clearly where it had wanted to park The same is true for cars 2 and 3 But when car 4 arrives, and parks at 4, maybe it wanted to park there, but since space 3 is occupied, it could very well have wanted to park at 3 Similarly car 5 may have wanted to park at 7 or 6, car 6 may have wanted to park at 2 or and car 7 may have wanted to park at 5, 4, 3, 2 or More generally, if α s = a and a,a 2,,a k are in the set {α,α 2,,α s } but, a k is not, let us call the string a, a,,a k the tail of α s otherwise, if that does not happen we will let α s be its own tail In any case we will refer to it as tail(α s,α) or briefly tail(α s ) This given, we see that the children of every node of the tree at level s are to be labeled from right to left by the tail of α s Moreover if we label the edges leading to the children of a node at level s, from right to left, by,q,q 2,,q k,then

25 Combinatorics of Tesler matrices 475 its is easy to see that the tree will have the further property that, as we go from the root down to a leaf, reading the labels of the successive nodes and multiplying the powers of q encountered in the successive edges, we will obtain the preference function corresponding to that leaf together with q to the power the extra travel that the cars end up traveling as a result of that preference Note that the total amount of travel the cars end up doing after they enter the one way street is ( ) n+ 2 (in parking space units) Since the sum of the components of the preference function gives the amount of travel they were hoping to do, the difference, (which is none other than the area of the corresponding, Dyck path) is, in fact, the extra travel they were forced to do This proves (5) Note that, for our example we also have the identity N (3,6,,4,7,2,5) (q) =[2] q [2] q [2] q [5] q Likewise, in the general case, denoting by #S the cardinality of a set S, we will have (52) N α (q) = n [#tail(α s )] q s= This proves our last assertion and completes the proof Levande proves that the same family of polynomials may be obtained from the subfamily of Tesler matrices in T [,,,] that have only one non-ero element in each row Since in the n n case this family has n! elements we will call these matrices Permutational Tesler orbriefly Permutational and denote them Π[,,,] Given a permutation α =(α,α 2,,α n ), it will be convenient to call the smallest element α j >α s, that is to the right of α s, the target of α s in α, and let α s be its own target if all the elements to its right are smaller that it In symbols { α s if max{α s+,,α n } <α s, target[α s,α]= min{α j : α j >α s &j>s} otherwise This given Levande s result may be stated as follows Theorem 52 For each n there is a bijection α A(α) between S n and Π[,,,]such that, if we set for each A = a i,j n i,j= (53) W Π [A] = [a i,j ] q a i,j>0

26 476 D Armstrong et al Then (54) W Π [A(α)] = N α (q) α S n Proof The construction of a matrix A(α) Π[,,,] satisfying (54) for each filling permutation α is immediate In fact, we simply fill row s by placing the integer #tail(α s ) in column j, ifthetargetofα s is α j Ofcourse, if α s is its own target then the integer #tail(α s ) ends up in the diagonal The reader will have no difficulty seeing that this recipe applied to the permutation α =(3, 6,, 4, 7, 2, 5) yields the permutational Tesler matrix on the right Since this construction is easily shown to be reversible, and therefore injective, and the cardinalities are in both cases n!, the only thing we need, to complete the proof, is to show that the resulting matrix is indeed in Π[,,,] as asserted This will be an immediate consequence of the following auxiliary fact Lemma 5 Suppose that for a given α =(α,α 2,,α n ) S n and some <k n we have Then {α s <α k : target[α s,α]=α k } = {α s,α s2,,α sr } (with s <s 2 < <s r <k) (55) tail(α k )={α k } + tail(α s )+tail(α s2 )+ + tail(α sr ) the + denoting disjoint union Proof Note that we must have α s >α s2 > >α sr, since if for some i we had α si <α si+ then the target of α si would be α si+ and not α k It follows from this that for each i we must have min tail(α si ) >α si+ In fact, with the opposite inequality we would have min tail(α si ) <α si+ <α si and since tail(α si ) is the interval of integers between its minimum and its maximum α si, this would force α si+ to be in tail(α si ) which contradicts

27 Combinatorics of Tesler matrices 477 s i <s i+ Now suppose that for some i we have α si+ < min tail(α si ), and set a =mintail(α si ) Clearly a must be to the right of α si in α, for otherwise a would in tail(α si ) But on the other hand a can t be on the right of α si+ for otherwise α si+ would not have α k as its target Now a cannot beitsowntargetsinceitislessthatα k and to the left of it Moreover a cannot have a k as its target, and thus its target, say b, must be less than α k However b must be greater than α si since all the elements in the interval [a +,α si ]areintail(α si ) Since b is is necessarily to the right of α si,this negates that α k is the target of α si This final contradiction proves that α si+ =mintail(α si ) This means that all the tails in (55) must be contiguous To prove (55) it remains to show that () the entry min tail(α sr ) is not on the left of α k and (2) 2 α s = α k Suppose that a = mintail(α sr ) istotheleftofα k, then, as in the previous argument a must be to the right of α sr since otherwise it would be part of tail(α sr ) But then as before the target b of a is shown to be a number greater that α sr,totherightofitandlessthanα k, and this negates that α k is the target of α si This contradiction proves () Next let a = α k and suppose that α s <abutthena can t be to the right of α s for the α k would not be the target of α s But if it is to the left of α s and therefore also to the left of α k,thenα k would be its target contradicting that α s,α s2,,α sr are all the element of α that are less than α k and have α k as a target This completes the proof of (55) To complete the proof of Theorem 52, recall that the integer we place in row k gives the sie of the tail of α k Thus the Tesler matrix condition for that row will be satisfied if and only if #tail(α k ) equals plus the sum of the integers we placed in column k, Now if these integers were placed in rows s <s 2 < <s r then this condition is simply (56) #tail(α k )=+a s,k + a s2,k + + a sr,k But by our construction, α s,α s2,,α sr must have been the entries of α with target α k, thus for each i r we must have set a si,k =#tail(α si ) Thus (56) is none other than #tail(α k )=+#tail(α s )+#tail(α s2 )+ +#tail(α sr ), which is clearly an immediate consequence of (55)

28 478 D Armstrong et al 6 Tesler matrices with arbitrary hook sums and weighted parking functions In (6) we set (6) P a,a 2,,a n (q, t) =( M )n A T (a,a 2,,a n) w H (A) with w H (A) = i,j h ai,j [ M] Now it is easy to show that we have for any integer k 0 if k<0 (62) h k [ M] = if k =0 M[k] qt if k>0 where (63) [k] qt = qk t k q t = q k + q k 2 q + + t k 2 q + t k Sinceaswehavenoted,whena,a 2,,a n are positive then all Tesler matrices in T (a,a 2,,a n ) have at least one non-ero element in each row, the identity in (62) assures that P a,a 2,,a n (q, t) is a polynomial In Section 4 we have seen a variety of cases where these polynomials are guaranteed to have positive integer coefficients Computer experimentations have revealed a variety of other cases In particular it is conjectured in [9] that the same thing happens when the a,a 2,,a n is any weakly increasing sequence In an effort to discover what collection of objects is q, t-enumerated by these polynomials, several values were obtained for q = t = The resulting data led the first author to conjecture (in a Banff meeting) that ( )( ) P a,a 2,,a n (, ) = a a + na 2 a + a 2 +(n )a 3 (64) (a + a a n +2a n ) In this section we give two proofs of this conjecture One uses an inductive argument and algebraic manipulations, while the other gives an explicit combinatorial interpretation of the conjecture by extending Levande s ideas As

29 Combinatorics of Tesler matrices 479 a by-product we will obtain a parking function interpretation for the matrices Π[a,a 2,,a n ] To see how this comes about we need some preliminary observations Note first that setting q = t = in (6) reduces it to a sum over Permutational Tesler matrices Now set (65) Q a,a 2,,a n (q, t) = w D (A) with (66) w D (A) = A Π(a,a 2,,a n) a i,j>0 [a i,j ] qt Then, the identities in (62) imply that for all positive choices of a,a 2,,a n we have (67) P a,a 2,,a n (, ) = Q a,a 2,,a n (, ) Keeping this in mind, for a given parking function PF PF n,letusset, for pf = pf(pf) (68) y(pf) = where the y i s are variables and the word π = π π 2 π n is obtained by combining the preference function pf corresponding to PF with its parking permutation More precisely, if pf =(pf[],pf[2],,pf[n]) and parking(pf) = (β,β 2,,β n )then n (69) π i = β pf[i] Note further that it makes perfectly good sense to consider the matrices Π[y,y 2,,y n ], In fact, for each of the n! choices of where to put the n non ero entries in the rows of a permutational matrix, by proceeding from top to bottom we can simply place in row s the sum of y s plus the entries in the previous rows that have been placed in column s Theentriesofamatrix A(y) = a i,j n ij= are of course linear forms in y,y 2,,y n and we will set as before (60) w ( A(y) ) = a i,j (y) y πi a i,j(y) 0 This given we have the following fact

30 480 D Armstrong et al Theorem 6 For each n there is a bijection α A(α, y) between S n and Π[y,y 2,,y n ] such that (6) y(pf) =w ( A(α, y) ) Moreover we also have (62) filling(pf)=α PF PF n y(pf)= A(y) Π[y,y 2,,y n] w ( A(α, y) ) = y ( y + ny 2 )( y + y 2 +(n )y 3 ) (y + y y n +2y n ) Proof The identity in (6) is another consequence of Lemma 5 To begin, for a given α S n and β = α we construct the matrix A(α, y) byplacing (for s =, 2,,n)inrows and column j the linear form (63) a s,j (y) = a tail(α s) if α j is the target of α s in α The same tree we used in the proof of Theorem 5 yields the identity y βa (64) n y(pf) = y πa filling(pf)=α s= a tail(α s) Thus to complete the proof of (6) we need only verify that the resulting matrix A(α, y) lies in Π[y,y 2,,y n ] For this we need that the linear form we place in row k equals y[β αk ]+ y[β a ]+ y[β a2 ]+ + y[β ar ] a tail(α s ) a 2 tail(α s2 ) a r tail(α sr ) where again α s,α s2,,α s3 are the entries of α whose target is α k But this is again a consequence of applying y[β] to every element of the identity in (55) This given, the first equality in (62) follows from (6) To prove the second equality we can follow two different paths, either algebraic manipulations or direct combinatorial counting We will do it both ways since each argument yields identities that may be useful in further work We start with a simple combinatorialiation of (64)

31 Combinatorics of Tesler matrices 48 Proposition 6 (65) y(pf)=y PF PF n n (y + y y i +(n +2 i)y i ) i=2 Proof The monomial y(pf) contributed by a parking function PF to the sum on the left-hand side of (65) may be progressively obtained by the following steps as the cars arrive one by one To begin, car contributes the factor y Having constructed the partial monomial y j y j2 y ji after the arrival of cars, 2,,i, the factor y ji contributed by car i is y i if the parking space pf[i] isfreeandisy r if space pf[i] is already occupied by car r there after car i proceeds to the first free space To show that the polynomial resulting by summing these monomials factors as in (65), we make use of the original device of parking the cars on a circular one way street with n + parking spaces Here again we construct each monomial as indicated above but from the unrestricted family of preference functions giving n + parking choices for each car We have then n + choices for car For car 2, its choice could be the place where car parked, in that case the partial monomial would be y 2 There remain n free places where car 2 can park, in which case the partial monomial is y y 2 Now two spaces are occupied, one by car and one by car 2 So when car three arrives the possible new factors it can contribute are y,y 2 and in n different ways y 3 We can clearly see now how the successive factors at the right-hand side of (65) do arise from this construction Finally, when car n comescars, 2,,n are already parked, and the final factor in the resulting monomial can be one of y,y 2,,y n and y n in two different ways It follows that, by summing the monomials produced by all these (n +) n preference functions, we must necessarily obtain the polynomial n (66) (n +) y (y + y y i +(n +2 i)y i ) i=2 Because of the rotational symmetry of this particular parking algorithm, it follows that for each of the n + parking spaces, by summing the monomials coming from the preference functions that leave that space empty, we must invariably produce the same polynomial Therefore by summing the monomials produced by the preference function that leave space n+ empty we will obtain the polynomial in (66) divided by n + This proves (65) and completes the proof of (62) as well

32 482 D Armstrong et al We give next an inductive proof of (62) Proposition 62 (67) w ( A(α, y) ) = y n A(y) Π[y,y 2,,y n] i=2 (y +y 2 + +y i +(n+2 i)y i ) Proof Note first that the left-hand side of (67) for n = 2 reduces to [ ] [ ] y 0 0 y w + w = y 0 y 2 0 y + y y 2 + y 2 + y y 2 = y (y +2y 2 ) 2 Thus (67) holds true in this case We can thus proceed by an induction argument, but for clarity and simplicity this idea is best understood in a special case We will carry it out for n = 5 and that will be sufficient Let us denote the polynomial produced by the left-hand side of (67) by L[y,y 2,,y n ] Note that it immediately follows from the definition of the matrices in Π[y,y 2,,y n ], by taking into account the column occupied by the nonero element of the first row, that we must have the following recursion for n =5 L[y,y 2,y 3,y 4,y 5 ]=y L[y 2,y 3,y 4,y 5 ]+y L[y + y 2,y 3,y 4,y 5 ] + y L[y 2,y + y 3,y 4,y 5 ]+y L[y 2,y 3,y + y 4,y 5 ] + y L[y 2,y 3,y 4,y + y 5 ] Using MAPLE, the inductive hypothesis gives that these five L-polynomials are as listed in the display below (68) y 2 (y 2 +4y 3 )(y 2 + y 3 +3y 4 )(y 2 + y 3 + y 4 +2y 5 ) (y + y 2 )(y + y 2 +4y 3 )(y + y 2 + y 3 +3y 4 )(y + y 2 + y 3 + y 4 +2y 5 ) y 2 (4y + y 2 +4y 3 )(y + y 2 + y 3 +3y 4 )(y + y 2 + y 3 + y 4 +2y 5 ) y 2 (y 2 +4y 3 )(3y + y 2 + y 3 +3y 4 )(y + y 2 + y 3 + y 4 +2y 5 ) y 2 (y 2 +4y 3 )(y 2 + y 3 +3y 4 )(2y + y 2 + y 3 + y 4 +2y 5 ) Notice that the three middle polynomials have a common factor of y + y 2 + y 3 + y 4 +2y 5, while the first and last start with the same three factors, while adding their last factors gives y 2 + y 3 + y 4 +2y 5 +2y + y 2 + y 3 + y 4 +2y 5 =2(y + y 2 + y 3 + y 4 +2y 5 )

33 Combinatorics of Tesler matrices 483 Thus omitting the common factor (y + y 2 + y 3 + y 4 +2y 5 ), combining the first and last polynomial in (68) and the remaining three polynomials we have (69) 2y 2 (y 2 +4y 3 )(y 2 + y 3 +3y 4 ) (y + y 2 )(y + y 2 +4y 3 )(y + y 2 + y 3 +3y 4 ) y 2 (4y + y 2 +4y 3 )(y + y 2 + y 3 +3y 4 ) y 2 (y 2 +4y 3 )(3y + y 2 + y 3 +3y 4 ) Now we see the same pattern emerging Summing the last factors of the first and last in (69) gives 2(y 2 + y 3 +3y 4 )+3y + y 2 + y 3 +3y 4 =3(y + y 2 + y 3 +3y 4 ) Omitting the common factor (y + y 2 + y 3 +3y 4 ) and combining the first and last now gives (620) 3y 2 (y 2 +4y 3 ) (y +y 2 )(y + y 2 +4y 3 ) y 2 (4y + y 2 +4y 3 ) Now adding the last factors of the first and last gives 3(y 2 +4y 3 )+4y + y 2 +4y 3 )=4(y + y 2 +4y 3 ) Omitting the factor y + y 2 +4y 3 and combining the first and last give 4y 2 y + y 2 whose sum is y +5y 2 which is the last factor we needed to complete the induction and the proof 7 The weighted parking function setting for q-teslers The problem that still remains open is to discover what combinatorial structures are q, t-counted by the polynomial in (6), namely (7) P u,u 2,,u n (q, t) =( M )n w H (A) A T (u,u 2,,u n)

34 484 D Armstrong et al where we replaced the a i s by the u i s to avoid notational conflicts Theorem 6, and more specifically the identity in (62), essentially gives a parking function setting for the polynomial P a,a 2,,a n (, ) In this section we go one step further towards the general open problem by working out the statistic that gives a parking function setting for the polynomial P u,u 2,,u n (q, ) Note first that setting t = in (7) reduces it to the polynomial (72) P u,u 2,,u n (q) = with (73) w q (A) = A Π(u,u 2,,u n) a i,j>0 [a i,j ] q w q (A) We will next define the q,u-weight of a parking function PF which we will denote w q,u [PF] To this end, let α = filling(pf)andβ = parking(pf) and pf be the corresponding preference function Now recall that for a given i n, α i gives the space that is occupied by car i By definition tail[α i ] gives the possible places that car i may have prefered to park Thus pf[i] must necessarily be an element of tail[α i ] More precisely, if tail[α i ]= [α i k i,,α i,α i ]thenpf[i] =α i r i, for some 0 r i k i Thisgiven, let (74) ex(α, r i )= r i+ j k i u βαi j and keeping in mind that, given that filling(pf) = α, the preference function pf as well as PF are completely determined by the sequence (r,r 2,,r n ), we set (75) w q,u [PF]= n q ex(α,ri)[ u βαi r i ]q We are now in position to state and prove Theorem 7 For every positive integral vector u =(u,u 2,,u n ) we have (76) w q,u (PF)=P u,u 2,,u n (q) PF PF n

35 Combinatorics of Tesler matrices 485 Proof Recall that a matrix A = a i,j n i,j= UPis in Π(u,u 2,,u n )if and only if there is one non-ero element in its row and if the element in row k is a k,j then given that the non-ero elements in column k are (77) a s,k,a s,k,,a sr,k, we must have (78) a k,j = u k + a s,k + a s,k + + a sr,k We will obtain all these matrices by a construction that follows closely the proof of Theorem 6 More precisely for a given α S n we will let A(α) be a permutational matrix A = a i,j n ij= UPobtained by placing, the non-ero element in row s, in column k if α k is the target of α s But here, to satisfy the requirement in (74), we must set (79) a s,k = a tail(α s) with β = α The reason for this is that if the non-ero elements in column k are a s,k,a s2,k,,a sr,k, then the identity in (55) gives a tail(α k) u βa u βa = u k + a s,k + a s2,k + + a sr,k which is (74) since the sum on the left-hand side is none other than that of the non-ero element we place in row k This identity gives that A(α) Π(u,u 2,,u n ) for all S n and since this construction is reversible thus injective and the cardinalities are the same, it must also be surjective Note next that it follows from our definition in (74) that for a fixed i n we have, for tail[α i ]=[α i k i,,α i,α i ] k i r i=0 q ex(α,ri)[ u βαi r i ] q =[u βαi k i ] q + q uβ α i k i [u βαi k i + ] q + q uβ α i k i +u βαi k i + [u β αi k i +2 ] q + + q uβ α i k i + +u βαi [u β αi ] q = [ a tail[α i] u β a ]q = [ a i,ji (α) ] q,

36 486 D Armstrong et al where the last equality follows from the definition in (79) provided that the non-ero element in row i of the matrix A(α) is in column j i Thus it follows from (75) that (70) filling(pf)=α w q,u [PF]= n ( ki r i=0 q ex(α,ri)[ u βαi r i ] q ) = w q [A(α)] and the equality in (76) is then obtained by summing over all α S n This completes our proof The special case t = /q of the bigraded Hilbert series for diagonal harmonics is known to have a nice formulation, which in the language of (6) can be expressed as q (n 2) P(,,,) (q, /q) =(+q + q q n ) n Maple calculations indicate that this nice special case extends in the following elegant way Conjecture 7 Let (a,a 2,,a n ) be an arbitrary sequence of positive integers, and set n F = (n + i)a i n Then q F P (a,a 2,,a n)(q, /q) =[a ] q n+ n i=2 [ i ] a j +(n +2 i)a i j= 8 The Dyck path associated to a Tesler matrix In this section we show how to associate a Dyck path to a Tesler matrix in T [,,,,], which leads to prime parking functions and refinements of some of our results and conjectures on Tesler matrices Let us say that an n n matrix A = a i,j n i,j= factors at k n if it is block diagonal with an initial block of sie k k More precisely, A factors at k if and only if a i,j =0when i k & j>kand when i>k& j k; in that case, the two submatrices A = a i,j k i,j= and A 2 = a i,j k<i,j n will be referred to as factors ofa It develops that a very simple factoriation criterion follows in full generality straight from the definition of Tesler matrices q

37 Combinatorics of Tesler matrices 487 Proposition 8 For any Tesler matrix A = a i,j n i,j= T[u,u 2,,u n ] we have the equality (8) k k n a s,s + a i,j χ(i <j)= k s= j=k+ s= u s (for all k n) Thus we have the inequality (82) k a s,s k s= s= u s with equality for k = n Moreover, we see that A factors at k if and only if (83) k k a s,s = u s s= s= Proof Note that the hook sum equalities for a Tesler matrix may be written in the form n n a s,j χ(j s) a i,s χ(i <s)=u s j= Summing for s k gives k a s,s + s= n j= s= k a s,j χ(j >s) n s= k a i,s χ(i <s)= k u s Replacing s by i in the first sum and s by j in the second sum gives k a s,s + s= k j= n a i,j χ(i <j) k j= k a i,j χ(i <j)= and (8) follows by canceling equal terms But now we can clearly see that (82) can hold if and only if k n j=k+ a i,j χ(i <j)=0 Thus the last assertion is a immediate consequence of the non-negativity of the entries of Tesler matrices s= k s= u s

38 488 D Armstrong et al The inequality in (82) allows us to associate to each matrix in T [,,,] a Dyck path by the following construction Given A = a i,j n i,j= T [,,,], starting at (0, 0) of the n n lattice square, and for each diagonal element a i,i, we go successively NORTH one unit step and then EAST a i,i unit steps The inequality in (82) (when all u i =) assures that the resulting path remains weakly above the diagonal (i, i) Moreover, since each time equality holds in (82), the factoriation criterion applies, and there is a direct connection between the Dyck path and the block diagonal structure of the corresponding matrix In particular it follows from this that the Tesler matrix does not factor at any k if and only if its Dyck path never hits the diagonal Computer calculations indicate that the weighted sum of these matrices yields a polynomial with positive coefficients, whose evaluation at t = q = appears to be enumerating structures equi-numerous with the prime parking functions of Novelli-Thibon [4] Further explorations with the Tesler matrices in T [,,,] led to a variety of other conjectures relating the shape of the Dyck paths to the combinatorial properties of the sum of their weights To state these findings we need some auxiliary facts and definitions To begin, given a preference function f =(f,f 2,,f n ), we will say that it factors at k<nif max{f i : i k} = k and min{f i : k + i n} = k + If that happens we say f itself may be expressed as a product by setting f = g h with g = (f,f 2,,f k )andh = (f k+ k, f k+2 k,,f n k) Since both g and h are necessarily preference functions for one way streets with k and n k spaces respectively, we are led in this manner to a unique factoriation of each preference function into a product of prime preference functions Denote by π n the number of prime preference functions f = (f,f 2,,f n ) Using the unique factoriation, together with the fact that (n +) n gives the total number of preference functions f =(f,f 2,,f n ), Novelli and Thibon derive that n π n t n = + n (n +)n t n (84) = t +2t 2 +t 3 +92t t t t 7 + O ( t 8) Recall that we have defined the diagonal composition of a Dyck path D and have denoted it p(d) the length of the intervals between successive diagonal hits Note that for a Dyck path in the n n lattice square that returns to the diagonal only at the end we will set p(d) =(n)