describe next. (In fact Chevalley [C] has shown a more general result probably earlier, but his paper was unpublished until recently.) For a given 2 S

Transcription

1 ON THE MULTIPLICATION OF SCHUBERT POLYNOMIALS RUDOLF WINKEL Abstract: Finding a combinatorial rule for the multiplication of Schubert polynomials is a long standing problem. In this paper we give a combinatorial proof of the Bergeron-Billey rule, which says how to multiply a Schubert polynomial by a complete or elementary symmetric polynomial, and describe some observations in the direction of a general rule. To every nite permutation contained in one of the symmetric groups S n there is associated a Schubert polynomial X 2 Z[x] Z[x ; x 2 ; x 3 ; : : : ], which is invariant under the natural embedding S n,! S n+ given by () : : : (n) 7! () : : : (n) (n+). We write S for the direct limit of all the S n under this inclusion so that e.g. = 23 = 234 = 2345 = : : : 2 S. The multiplication of Schubert polynomials is well known to represent faithfully the `Schubert calculus' of cohomology classes of Schubert varieties of ag manifolds with respect to the cup-product, i.e. intersection. Since the set of all Schubert polynomials forms a Z-basis of the algebra Z[x] this geometric interpretation shows that the structure constants c 00 occurring 0 in (0.) X X 0 = X 00 c 00 0X 00 are all non-negative integers. This suggests the possibility of a combinatorial rule for the determination of these constants, but over the years only little progress has been made in solving this problem, which quite aptly can be called the most important open problem in the \elementary" theory of Schubert polynomials. (It is of course possible to compute the product X X 0 explicitly and expand it afterwards using the property that every Schubert polynomial has the form X = x L() + : : :, where the Lehmer code L() of (see below) is the smallest exponent with respect to the lexicographic order induced by 0 < < 2 < : : :.) The existence of Schubert polynomials and many of their properties have been established in a sequence of papers by A. Borel (953), I.N. Bernstein, I.M. Gelfand, and S.I. Gelfand (973), M. Demazure (973-74), and nally A. Lascoux and M.-P. Schutzenberger (982-87). For comprehensive accounts on the geometry and combinatorics of Schubert polynomials see [Hi] and [LS, M, M2, W]. We expose here only those parts of the theory, which are strictly necessary for our presentation. For k < l let kl (k; l) be the transposition interchanging k and l, and k := (k; k+). The complete and elementary symmetric functions in m variables of degree N are denoted by h (m) N = h N (x ; : : : ; x m ) and e (m) N = e N(x ; : : : ; x m ), respectively. The Schubert polynomial X k is then given by X k = x + + x k = h (k) = e (k). Monk has established in [Mo] a combinatorial rule for the multiplication of a Schubert polynomial by a X k, which we will Date: Oktober Mathematics Subject Classication. 4M5, 05E5. Supported by Deutsche Forschungsgemeinschaft (DFG).

2 describe next. (In fact Chevalley [C] has shown a more general result probably earlier, but his paper was unpublished until recently.) For a given 2 S and k < l we write (0.2) (k) / (l) :() (k) < (l) and ]f j k < < l; (k) < () < (l)g = 0 : It is well known (see e.g. [W, Prop.2.3]) that (0.3) l( kl ) = l() + () (k) / (l) ; where l() is the length of the permutation. The notation (0.4) is helpful in dening the (; b)-admissible set: (k) b / (l) :() (k) / (l) and k b < l (0.5) J(; b) := f(k; l) j (k) b / (l)g : Theorem 0.. (Monk) With the above notations one has for every 2 S and b 2 N : (0.6) X b X = X (k;l) 2J(;b) X (k;l) : Simple proofs can be found in [M2, W]. Notice that `/' is the covering relation for the Bruhat order on S and ` b/' for the b-bruhat order, so that Monk's rule can be rephrased by saying: \The product of X with X b is the sum over all Schubert polynomials indexed by all permutations covering in b-bruhat order." Example 0.2. Let = 2543 and b = 3; in other words; we want to compute X 3 X Since 3542; 26435; and cover 2543 in 3-Bruhat order the result is X X X Only recently N. Bergeron and S. Billey [BB] have, unaware of the fact that A. Lascoux and M.-P. Schutzenberger had stated an equivalent rule in a dierent form in [LS2], formulated a conjecture (described in Section below), which extends the Monk's rule from the case h (m) = e (m) to the cases of arbitrary complete and elementary symmetric polynomials h (m) N and e (m) N. These conjectures have been proven by F. Sottile [So] using an explicit geometric description of certain intersections of Schubert varieties. We will instead give in Section a combinatorial proof of these facts departing directly from Monk's rule and using only manipulations of permutations. Despite its conceptual simplicity the checking of the details is quite tedious and tricky { an appearend feature already of Sottile's geometric proof. In the subsequent sections we describe some observations and \approximative rules" of increasing generality, how to multiply a Schubert polynomial by an Schur polynomial (Sec.2), by a Schubert polynomial associated to a L-unimodal permutation (Sec.3), and by a general Schubert polynomial (Sec.4). We speak of \approximative rules", because these observations give the correct results for many examples, but are not as yet reliable rules or even steady conjectures. The appearent diculty of the problem may serve as an excuse for including this increasingly more speculative material. The `L' in L-unimodal stands for the Lehmer code L() of a permutation 2 S n, i.e. L() 2 f l n? ; : : : ; l 0 j 0 l n? n? ; = ; : : : ; n g ; 2

3 where l n? () := ]f j j < j; () > (j)g for all 2 f; : : : ; ng, e.g. L(36542) = or L(257346) = It is not hard to see that can be reconstructed from its Lehmer code so that L can be regarded as a bijection between permutations and codes. A permutation is called Grassmannian i there is a partition : : : s ( : : : s ) and a natural number m l() = s such that L() = 0 : : : 0 s : : : 0 : : : 0 with m? s 0 zeros on the left and (at least) zeros on the right. An alternative denition is: is called Grassmannian i has a at most one descent, i.e. there is at most one i with (i) > (i + ). Anyway (0.7) Then (; m) := L? (0 : : : 0 s : : : {z } m 0 : : : 0) : (0.8) X (;m) = s (m) (x) ; in other words: a Schubert polynomial X is a Schur polynomial exactly when is Grassmannian. A permutation is called dominant i its Lehmer code is non-increasing; then X = x L() = x l n?() x l n?2() 2 : : :. (For proofs of the above facts see [M2] or [W]). Obviously both Grassmannian and dominant permutations are special cases of L-unimodal permutations, which are permutations having an unimodal Lehmer code: l n? l n? l 0 for some. The general case, where the permutation is not necessarily L-unimodal, is treated in Section 4. Hereby and already in Sections 2 and 3 we rely strongly upon the notion of the diagram D() of a permutation 2 S n, which is a subset of an n n-array of unit squares or boxes in the plane: D() f[i; j] 2 Z Z j i; j ng such that D() originates from f[i; j] j i; j ng by cancelation of the `hooks' of boxes (i = ; : : : ; n): For example = has the diagram f[i 0 ; (i)] j i 0 ig [ f[i; j 0 ] j j 0 (i)g : , where we have added dots in the positions (i; (i)) indicating the `corners' of the hooks removed. We will use the notation (i; j) for the position `column i, row j' and [i; j] for the box in position (i; j). Note that the number of boxes in column i of D() is equal to the i-th entry of L(). Subsequently we will always use the convention that empty rows are removed from the diagram D() and we reduce the n n frame box to some lines, which indicate the left border and the relative positions of boxes; e.g. D(26354) = 3

4 Moreover the observations in Sec.3 and 4 will be based on the notion of a component of a diagram D(): two boxes in D() are said to be adjacent, if they share a common edge in D() (without empty rows!); then a component of D() is dened as an element of the partition of boxes of D() induced by adjacency as equivalence relation. It is not hard to see (cf. Lem.3.) that D() contains exactly one component i is L-unimodal; and in Sec.4 we will argue that the `interaction' of several components in D() is the cause, which makes it dicult to nd a simple combinatorial rule for the multiplication of Schubert polynomials. Finally we remark that the examples in the present paper (and many more not included here) are checked with the help of the package symmetrica maintained at the University of Bayreuth, Germany.. The Bergeron-Billey rule In [BB] N. Bergeron and S. Billey conjectured a combinatorial rule, which says how to multiply a Schubert polynomial by a complete symmetric or elementary symmetric polynomial. For the description and proof of this rule we have to extend the notations (0.2-5) given in the introduction: (.) (.2) (.3) (.4) (.5) (.6) X m :=X m ; (k; l; N) :=((k ; l ); : : : ; (k N ; l N )) with k i < l i ; i = ; : : : ; N ; (i) (i) (k; l; N) :=(k ; l ) : : : (k i ; l i ) for given 2 S ; 0 i N ( (0) := ); where the `-specic condition' is (.7) (.8) :=( ; : : : ; N ) with i 2 N ; i = ; : : : ; N ; J(; ) :=f(k; l; N) j (k i ; l i ) 2 J( (i?) ; i ) for i = ; : : : ; Ng ; J(; ; ) :=f(k; l; N) 2 J(; ) j -specic condition g ; (i) (k i ) < (i+) (k i+ ) for i = ; : : : ; N? ; if = N ; (i) (k i ) > (i+) (k i+ ) ; for i = ; : : : ; N? ; if = N : (.) is a abbreviation, (.2) the notation for an arbitrary sequence of of length N of transpositions, (.3-4) select more specic sequences in the spirit of (0.4), and the sets J(; ) and J(; ; ) of (.5-6) are of course called the (; )- and (; ; )-admissible sets in generalization of (0.5). Theorem.. Let 2 S, m N 2 N, = ( ; : : : ; N ) = (m; : : : ; m) m N, and 2 fn; N g. Then with the above notations: (.9) X (;m) X = X (k;l;n )2J(;;) X (N) : Proof. We concentrate on the `complete symmetric' case, because the `elementary symmetric' case is analogous using (.8) instead of (.7); alternatively and more conveniently one can argue with `duality' as in [So, Lem.2]. For h (m) N = h N(x ; : : : ; x m ) = X (N;m) one has the simple recursion (.0) h (m) N = h(m?) N + x m h (m) 4 = N? h(m?) N + (X m? X m? )h (m) N? ;

5 which is of course equivalent to (.) X m? h (m)? N? h(m?) N = X m h (m)? N? h(m) N : The proof now proceeds by simultaneous induction over m and N: for m 2 N and N = equation (.9) is Monk's rule and for m = and N 2 N one computes by repeated application of Monk's rule x N X = x N? X (;l ) = : : : = X (;l ):::(;l N ) ; where l ; : : : ; l N is a unique sequence of natural numbers (depending on ) with the property 2 l < < l N. But because (i) (k i ) = (i) () / (i) (l i ) = (i+) () = (i+) (k i+ ) one sees that the N-tuple ((; l ); : : : ; (; l N )) is the single element of J(N; ; N ), which establishes (.9) in this case. We assume now that (.9) is true for the multiplication by h (m) and h(m?) N? N and use the notation (.2) (k N ; l N ) b * J(N? ; ; m N? ) to indicate the concatenation of the pair (k N ; l N ) with the property (N?) (k N ) b / (N?) (l N ) to the (N? )-tuples (k; l; N? ) 2 J(N? ; ; m N? ). A useful and necessary notation is (.3) S := f (N ) j (k; l; N) 2 Sg for S = J(; ); J(; ; ); etc.. for the multiset of `nal' permutations (N ) originating from some by a sequence of N transpositions. The main reason for introducing this notation is that we have to speak of a multiset of `nal' permutations independantly of the chain of steps through which these are reached. Taking into account the induction hypothesis, Monk's rule, equation (.), and Lemma.2 below we are left with the proof the following equality of sets: (.4) f(k; l; N) j (k N ; l N ) m? * J(N? ; ; m N? )g n J(N; ; (m? ) N ) = f(p; q; N) j (p N ; q N ) m * J(N? ; ; m N? )g n J(N; ; m N ) : Note that we need Lemma.2, because (.) contains negative summands, but we have no `negative sets'. The proof of Lemma.2 shows that the r.h.s. equals f(p; q; N) j (p N ; q N ) m * J(N? ; ; m N? ); (N?) (k N? ) = (N?) (p N? ) > (N?) (q N )g ; where the inequality is strict, because k N? m < q N. Since the two sets in (.4) are dierent only by virtue of the step N, we can collect our knowledge for the two pairs (p N ; q N ) and (k N ; l N ) as follows: (.5) (.6) r.h.s. (.4): (N?) (p N ) m / (N?) (q N ) (N?) (p N? ) > (N ) (p N ) l.h.s. (.4): (N?) (k N ) m? / (N?) (l N ) (N?) (k N? ) (N ) (k N ) ; where the last assertion follows, because we remove permutations related to the condition (N?) (k N? ) < (N ) (k N ). 5

6 Lemma.2. With the notations of the Theorem one has: J(N; ; (m? ) N ) f(k; l; N) j (k N ; l N ) m? * J(N? ; ; m N? )g ; J(N; ; m N ) f(p; q; N) j (p N ; q N ) m * J(N? ; ; m N? )g : Proof. Let rst (k; l; N) 2 J(N; ; (m? ) N ); if l ; : : : ; l N? 6= m, then (k; l; N? ) 2 J(N? ; ; m N? ) and consequently (k; l; N) 2 f(k N ; l N ) m? * J(N?; ; m N? )g. Assume now that l i = m for one i 2 f; : : : ; N? g; then (i) (k i ) = (i?) (l i ) = (i?) (m) > (i?) (k i ) = (i) (l i ) = (i) (m) and by the -specic condition one concludes that from (i) to (N ) 2 J(N; ; (m?) N ) there occurs no change on the places m and k i : This shows l i+ ; : : : ; l N > m and and consequently (i) (m) = = (N ) (m) and (i) (k i ) = = (N ) (k i ) : ((k ; l ); : : : ; (k i? ; l i? ); (k i+ ; l i+ ); : : : ; (k N ; l N )) 2 J(N? ; ; m N? ) ((k ; l ); : : : ; (k i? ; l i? ); (k i+ ; l i+ ); : : : ; (k N ; l N ); (k i ; l i )) 2 f(k i ; l i ) m? * J(N? ; ; m N? )g : But by the forgoing we have = (k ; l ) : : : (k i? ; l i? )(k i+ ; l i+ ) : : : (k N ; l N )(k i ; l i ) as desired. For the proof of the second inclusion observe that the dierence between the two sets is simply that for J(N; ; m N ) there is an additional requirement on the pair (p N ; q N ) namely by the -specic condition: (N?) (p N? ) < (N ) (p N ) = (N?) (q N ). We investigate conditions (.5) and (.6) more closely: in case of p N ; l N 6= m (and (N?) (k N? ) > (N ) (k N )) the two conditions are clearly equivalent (observe in particular that (N?) (k N? ) > (N?) (l N ), because k N? m < l N ). Therefore the problematic cases are l N = m (together with (N?) (k N? ) = (N ) (k N ) ) and p N = m. The remaining proof has two parts: we show rst that every (N ), which obeys (.6) and l N = m, can be found in the set of all (N ), which obey (.5) and p N = m; and second that the reverse is true. In fact the arguments below reveal that the two \critical cases" in (.5) and (.6) are the same: \l N = m () p N = m". For both inclusions the overall tactic is rst to show that a (N ) belonging to one critical case has special properties, which allow it to be reached through a chain of transpositions belonging to the other critical case, and second one has to check that the other path is indeed admissible in terms of the other set. Assume that (.6) and l N = m are valid for some (N ) ; then Lem..3 below shows that k N? = m. Lemma.3. With the notations of the Theorem one has for (N ) 2 f(k; l; N) j (k N ; l N ) m? * J(N? ; ; m N? )g with l N = m, that k N? = m. Proof. Let k N? < m and i := maxfj j j N? 2; k j = mg, otherwise we would have (N ) 2 J(N; ; (m? ) N ). To facilitate notation we assume i = so that (N ) = 6

7 (m; l )(k 2 ; l 2 ) : : : (k N? ; l N? )(k N ; m), where k 2 ; : : : ; k N < m and l ; l 2 ; : : : ; l N? > m, the latter being all distinct by the -condition. We then show (m; l )(k 2 ; l 2 ) : : : (k N? ; l N? )(k N ; m) () = (m; l )(k N ; m)(k 2 ; l 2 ) : : : (k N? ; l N? ) = (k N ; m)(k N ; l )(k 2 ; l 2 ) : : : (k N? ; l N? ) (2) 2 J(N; ; (m? ) N ) ; which proves k N? = m. () is valid, if k 2 ; : : : ; k N? 6= k N. Assume that := maxfj j 2 j N? ; k j = k N g. Then (N?) (m) = () (m) = () (k ) < (j) (k j ) = (j) (k N ) = (N?) (k N ) ; which is in contradiction to (N?) (k N ) / (N?) (m). For (2) we have to check that (a) (k N ) / (m) ; (b) (k N ; m)(k N ) / (k N ; m)(l ) ; (c) (k N ; m)(k N ) < (k N ; m)(k N ; l )(k N ) ; where (c) comes from the -condition. Is easy to see that (b) is equivalent to (m) / (l ), and similarly one computes for (c): For (a) observe that (k N ; m)(k N ) = (m) < (l ) = (k N ; m)(l ) = (k N ; m)(k N ; l )(k N ) : (k N ) / (m) / (l ) =) (b) : But it is easy to see that the negation of (k N ) / (m) together with (m) / (l ) implies the negation of (b); since (m) / (l ) and (b) are already seen to be true, (k N ) / (m) must be true. Since k N? = m, there must exist a j 2 N such that m = k N? = k N?2 = = k N?j > k N?j?. Note once more that l N? ; l N?2 ; : : : ; l N?j > m are all distinct by the -condition. Now with (.7) we dene (m; l N?j ) : : : (m; l N? )(k N ; m) = (k N ; l N?j ) : : : (k N ; l N? )(m; l N?j ) (p; q; N) : (k ; l ) : : : (k N?j? ; l N?j? )(k N ; l N?j ) : : : (k N ; l N? )(m; l N?j ) ; which says especially that p N = m! We have to show that (N ) (p; q; N) 2 f(p; q; N) j (p N ; q N ) m * J(N? ; ; m N? )g n J(N; ; m N ) : Let the (i) 's be associated to (k; l; N) and the e (i) 's be associated to (p; q; N), then obviously (i) = e (i) for i = ; : : : ; N? j? and i = N, and e (i) (p i ) < e (i+) (p i+ ) for i = ; : : : ; N?. It remains to be shown that () (N?j?) (k N ) / (N?j?) (l N?j ) ; (2) e (N?) (m) / e (N?) (q N ) ; (3) e (N?) (p N? ) > e (N?) (q N ) ; and (4) (N?j?) (m) / (N?j?) (k N ) ; 7

8 where (4) is auxiliary for (2), () is necessary for having (N?) (p; q; N? ) 2 J(N? ; ; m N? ) and (2), (3) for condition (.5). To show (4) assume rst that (N?j?) (m) > (N?j?) (k N ) and let := ]f j k N < < m; (N?j?) (k N ) < (N?j?) () < (N?j?) (m)g : In [W, Prop.2.3] it has been shown that under this circumstances interchanging m and k N increases the length of (N?j?) by + : l( (N?j?) (k N ; m)) = l( (N?j?) ) + +. We conclude therefore l(e (N?j) ) = l( (N?j) ) + + =) l(e (N?) ) = l( (N?) ) + j + j =) l(e (N ) ) l( (N ) ) + j + ; because during the j steps from (N?j?) to (N?) at most j? numbers greater (N?j?) (k N ) and less than (N?j?) (m) may have \immigrated" between the places k N and m ( at least e (N?) (l N? ) = (N?j?) (l N? ) is greater than (N?j?) (m) ). But l(e (N ) ) l( (N ) ) + j + is in contradiction to e (N ) = (N ). If on the other hand (N?j?) (m) < (N?j?) (k N ), then () follows, because the numbers on the places between k N and m remain unchanged for (N?j) ; : : : ; (N?), whence it would be a contradiction to (N?) (k N ) / (N?) (m), if () were not valid. () implies now (2), since e (N?) (m) = (N?j?) (m) / (N?j?) (k N ) = e (N?j) (l N?j ) = e (N?) (l N?j ) : For (3) one computes similarly that e (N ) (k N ) = e (N?) (k N ) = e (N?) (p N? ) > e (N ) (p N ) = e (N ) (l N?j ) () (N?j?) (l N? ) = (N?2) (l N? ) = (N?) (m) = (N ) (k N ) > (N ) (l N?j ) = (N?j) (l N?j ) = (N?j?) (m) () For () assume that there is with k N < < m so that (N?j?) (l N? ) (N?j?) (l N?j ). (N?j?) (m) : e (N?j?) (k N ) < e (N?j?) () < e (N?j?) (l N?j ) : Then we have a contradiction, because e (N?j?) (k N ) = e (N?) (k N ) and e (N?j?) (l N?j ) = e (N?j) (k N?j ) (:7) = e (N?) (k N? ) = e (N?) (m) : So far we have proved the inclusion `l.h.s. (.4)' `r.h.s. (.4)'. For the reverse inclusion we have to show that every (N ) 2 `r.h.s. (.4)', which obeys (.5) and p N = m, can be found in the set `l.h.s. (.4)' and obeys (.6) and l N = m. Note rst that p N = m implies p N? < m, because otherwise e (N?) (p N? ) = e (N?) (m) / e (N?) (q N? ) = e (N ) (p N? ) = e (N ) (p N ) ; in contradiction to (.5). Hence p N = m implies the existence of a j 2 N, such that (p N?j? 6= ) p N?j = = p N? < m. Lemma.4 below shows that we furthermore may 8

9 assume q N?j = q N. Therefore (.8) (p; q; N) =(p ; q ) : : : (p N?j? ; q N?j? )(p N? ; q N?j ) : : : (p N? ; q N? )(m; q N ) =(p ; q ) : : : (p N?j? ; q N?j? )(p N? ; q N?j ) : : : (p N? ; q N? )(m; q N?j ) =(p ; q ) : : : (p N?j? ; q N?j? )(m; q N?j ) : : : (m; q N? )(p N? ; m) (k ; l ) : : : (k N?j? ; l N?j? )(m; l N?j ) : : : (m; l N? )(k N ; m) ; i.e. l N = m! Lemma.4. With the notations of the Theorem one can for e (N ) (p; q; N) 2 f(p; q; N) j (p N ; q N ) * m J(N? ; ; m N? )g with p N = m assume without loss of generality that q N?j = q N. Proof. Assume that there is no i with i N? and q i = q N. If in addition all p i < m, then the transposition (m; q N ) in (.8) can be commuted in such a way that the -condition (.7) is fullled for all i; in other words: e (N ) 2 J(N; ; m N ). If on the other hand p i = m for some maximally chosen i N?, then by (.5) e (i) (p i ) = e (i) (m) = e (N?) (m) / e (N?) (q N ) = e (N ) (p N ) ( < e (N?) (p N? ) ) ; so that (m; q N ) can be commuted to one of the places i+; : : : ; N? with (.7) again fullled for all i. Therefore we conclude that q i = q N for some maximally chosen i N?. We rst investigate the possibility i = N? j? (the cases i < N? j? are similar). Let q N?j? = q N, then and e (N ) (m) = e (N?) (q N ) = e (N?j?) (q N ) = e (N?j?) (q N?j? ) < e (N?j?) (p N?j? ) ; (p N?j? ; q N )(p N? ; q N?j ) : : : (p N? ; q N? )(m; q N ) = (p N?j? ; q N )(m; q N )(p N? ; q N?j ) : : : (p N? ; q N? ) = (m; q N )(p N?j? ; m)(p N? ; q N?j ) : : : (p N? ; q N? ) = (m; q N )(p N? ; q N?j ) : : : (p N? ; q N? )(p N?j? ; m) : But the latter permutation with (p ; q ) : : : (p N?j?2 ; q N?j?2 ) in front is already an element of l.h.s. (.5), because the -condition (.7) is fullled for all i N? 2: e (N?j) (p N? ) > e (N?j?) (p N?j? ). e (N?j?) (q N?j? ) = e (N?j?) (m; q N )(m): The above arguments for the case i = N? j? can be carried out in exactly the same way for i = N? j (as representative for N? j i N? ) up to the point (p N? ; q N?j )(p N? ; q N?j+ ) : : : (p N? ; q N? )(m; q N ) = = (m; q N )(p N? ; m)(p N? ; q N?j+ ) : : : (p N? ; q N? ) ; but now the transposition (p N? ; m) can not be commuted to the right, whence without loss of generality only the case q N?j = q N remains to be considered. 9

10 In order to identify the permutation (N ) (k; l; N) of r.h.s. (.8) as an element of the set l.h.s. (.4) fullling (.6) we have to check that () e (N?j?) (m) / e (N?j?) (q N?j ) ; (2) (N?) (k N ) / (N?) (m) ; and (3) (N?) (k N? ) (N ) (k N ) : For (3) simply observe (N?) (k N? ) = (N?) (m) = (N ) (k N ). For () one notes that e (N?) (m)/e (N?) (q N ) implies that there is no with m < < q N?j such that e (N?j) (m) = e (N?) (m) < e (N?j) () < e (N?) (q N ) = e (N?) (q N?j ) = e (N?j) (q N?j ) ; because the numbers on places with m < < q N are not changed by the transpositions (p N? ; q N?j ); : : : ; (p N? ; q N? ). Therefore we can conclude that e (N?j?) (m) = e (N?j) (m) < e (N?j) (q N?j ) = e (N?j?) (p N?j ) = e (N?j?) (p N? ) ; and that there is no with m < < q N?j such that e (N?j?) (m) < e (N?j?) () < e (N?j?) (p N? ). This is exactly what is needed to conclude () from e (N?j?) (p N? ) / e (N?j?) (q N?j ). With regard to (2) it is easy to see by (.5) that (N?) (k N ) < (N?) (m), because (N?) (k N ) = (N ) (m) = e (N ) (m) = e (N ) (p N ) and (N?) (m) = (N ) (k N ) = e (N ) (p N? ) = e (N?) (p N? ). But for `/' one has to work a bit harder: note rst that (2) () e (N ) (m) / e (N ) (p N? ) ; i.e. e (N ) (p N? ) > e (N ) (m) and in addition for each with p N? < < m either e (N ) (m) > e (N ) () or e (N ) () > e (N ) (p N? ). But e (N?j?) (p N? ) / e (N?j?) (q N?j ) so that for each with p N? < m either e (N?j?) (p N? ) > e (N?j?) () or e (N?j?) () > e (N?j?) (q N?j ). Since moreover e (N ) (m) = e (N?) (q N ) = = e (N?j) (q N ) = e (N?j) (q N?j ) = e (N?j?) (p N? ) ; and e (N ) (p N? ) = e (N?) (p N? ) > > e (N?j) (p N? ) = e (N?j?) (q N?j ) by the -condition, we conclude e (N ) (m) = e (N?j?) (p N? ) / e (N?j?) (q N?j ) < e (N ) (p N? ) : But the numbers on places with p N? < < m are not changed by the transpositions (p N? ; q N?j ); : : : ; (p N? ; q N? ); (m; q N ), whence the desired conclusion is valid. This nishes the proof of the equality of sets (.4) and therefore of Theorem.. Example.5. We compute h (3) 3 X It is helpful to mark the position between places 3 and 4 by a vertical line, since this is the `axis' around which the transposition of numbers takes place. In addition we underline the numbers (i) (k i ), so that one can easily check the -condition (.7). 0

11 step: j3 354j2 364j j j35 274j j34 257j Of course the horizontal lines in the above table are to be understood as: 2543 in the rst step gives rise to 3542, 26435, and 25634; and 3542 in the second step splits up into and 35624, etc.. 2. Observations for the Grassmannian case Our overall assumption is that the multiplication of an arbitrary Schubert polynomial, say X 0, by another, say X, can be achieved through a sequence of l() steps of the `Monk form' (0.4-5) leading upwards in Bruhat order from 0, where the admissible, i.e. correct, paths are selected by a rule depending essentially only on the diagram D(). More precisely: the summands X (N), respectively the permutations (N ), in the product X 0X are reached by executing N = l() transpositions (k i ; l i ), i = ; : : : ; N, with (0) := 0 and recursively (i) := (i?) (k i ; l i ), where for each i there is a natural number i such that (i?) (k i ) i / (i?) (l i ). The number i will be called the (transposition) axis of the step i, and will in concrete calculations be represented as a vertical bar in (i?) between the places i and i +. The combinatorial rule, which says how to select an admissible transpositions (k i ; l i ) and an admissible axis i in each step (or admissible sequences (.2) and (.4)) will depend on the diagram D() (with empty rows removed). The rule uses the notion of the step tableaux S() for, which is a certain lling of D() with each of the numbers ; : : : ; N occurring once, and the notion of the weight tableaux W ( (N ) ) for each (N ), which is a lling of D() with the weights W (i) W ( (N ) ; i) := (i) (k i ) on the places i determined by the step tableaux S(). In the special case of Ex..5 one has = 3, h (3) 3 = X (3;3) = X 26345, = 2 = 3 = 3, and D(26345) =, S(26345) = 3 2, W (374256) = W (357246) = 7 6 3, and W ( ) = W ( ) =

12 More generally one can rephrase Theorem. as follows: Let 0 2 S, 2 fn; N g, m N, and = ( ; : : : ; N ) = (m; : : : ; m) m N. Then the admissible permutations (N ) (or admissible paths leading to some (N ) ) for the product X 0X (;m) are the following: If = N, then the entries ; : : : ; N are increasing from bottom to top in S((; m)), and so do the entries of each W ( (N ) ). If = N, then the entries ; : : : ; N are increasing from right to left in S((; m)), and the entries of each W ( (N ) ) are decreasing from right to left. In other words: the question, which (N ) are admissible, is reduced to examine the tableaux W ( (N ) ) in these cases, and the `admissible tableaux' W ( (N ) ) are described by the rule above. Below we describe an extensions of this notion of `admissibility of a weight tableaux' to the cases of Grassmannian permutations, which gives the correct result in many cases, but fails for example for X X { a simple counterexample found by N. Bergeron. Observation 2.. Recall the denition (0.7) of the Grassmannian permutation (; m) associated to a partition : : : s ( : : : s ) and a natural number m l() = s, and the fact that the number of boxes in column i of D() is equal to the i-th entry of L(). (Therefore D() is the Ferrer shape of with rows reected into columns as indicated by (0.7).) Then the product X 0X can be computed in many cases according to the following prescriptions: The step tableaux S() is given by numbering consecutively the columns from right to left and inside columns from bottom to top with the numbers ; : : : ; N = jj. = = N := m. W ( (N ) ) is admissible i the weights in W ( (N ) ) increase in every column from bottom to top and for i = ; : : : ; s? (2.) b j a j for j = ; : : : ; i+ ; where a ; : : : ; a i and b ; : : : ; b i+ are the entries of columns (m? i + ) and (m? i), respectively, where the indices increase from top to bottom. Note that condition (2.) can be rephrased by saying: if every column in a weight tableaux W ( (N ) ) is ushed to a common top line, then the the weights increase in every column from bottom to top and decrease in every row from right to left. Furthermore we remark that dierent `nal' permutations (N ) may have the same weight tableaux, but equal permutations (N ) necessarily have dierent weight tableaux reecting the fact that they are reached through dierent admissible path. Example 2.2. We compute X X = X s (4) 2 2 (x) with = = 5 = 4, D(35624) =, and S(35624) =

13 step: j4 2735j j j j j j j j j j3 2647j j j j j j j j j j j j j j j j346 Note that e.g. W (375624) = and W ( ) = Of course the Bergeron-Billey rule and the famous Littlewood-Richardson rule are contained as special cases in the above observation. Moreover the case of being a hook shape, proved as a simple consequence of the Bergeron-Billey rule in [So, Thm.8], is contained. 3. Observations for the L-unimodal case In the introduction we mentioned already that L-unimodal permutations generalize both Grassmannian and dominant permutations. In order to formulate Observation 3.2 for the multiplication of a Schubert polynomial by an X with L-unimodal we need the following simple Lemma 3.. a) is L-unimodal exactly when D() has one component. b) If is L-unimodal, then row j of D() includes row j + for all j. If the rightmost box in row j stands in column c j c j (), then one has c c 2 0. Proof. Let 2 S n be L-unimodal with l n? l n? l 0 for some. Then a) and b) are simple consequences of the generation rules for D(), namely the removal of the `hooks' of boxes h (i) := f[i 0 ; (i)] j i 0 ig [ f[i; j 0 ] j j 0 (i)g from the n n-array of boxes. Observe that if i <, then the removal of the h (i)'s generates the initial step-backs of 3

14 D(), and if < i n, then the removal of the h (i)'s generates proper inequalities in the sequence c c 2 : : :. Observation 3.2. Let 2 S n be L-unimodal as in the above lemma, and in particular let again c j be the column, which contains the rightmost box in row j, and r ; r 2 ; : : : ; r q be the rows for which c rj c rj +. Then exactly the (N ) occurring in the product X 0X are computed but possibly with a greater multiplicity (!) according to the following prescriptions: Subdivide D() into q subdiagrams of Grassmannian shape, i.e. the q subdiagrams G ; G 2 up to G q consisting of rows ; : : : ; r, rows r + ; : : : ; r 2, up to rows r q? + ; : : : ; r q, respectively. For p = ; : : : ; q the single subdiagrams of Grassmannian shape G p are treated now exactly as in Obs.2. or more correctly: \its yet to be found correct version", with m replaced by c rp, whereby one starts with G and works up to G q. For p = ; : : : ; q and g p := jg j + + jg p j we require that (gp) (k gp ) is not greater than any weight in the next higher row r p +. Example 3.3. For = S 9 diagram one has the Lehmer code L() = , the D() =, c = 7, c 2 = c 3 = c 4 = 5, c 5 = 4, and c 6 = 0, whence q = 3, r =, r 2 = 4, r 3 = 5. The subdiagrams of Grassmannian shape are G =, G 2 =, G 3 =, and nally S() = with g = 6, g 2 = 4, and g 3 = 5. Example 3.4. We compute X 3542 X 243 with = 243, L() = 20, = 3, 2 = 2, D() =, and S() = In the tableaux below we use the already established conventions, but add this time the `dead ends' of the computation with the respective nal steps in parenthesis. 4

15 step: j42 36j425 46j325 43j j (326j45) (346j25) 45j32 (435j2) (425j3) (345j2) (325j4) 3 If one tries a \column-wise" approach with step tableaux S() = 4 2 all reasonable combinations of 2 and 3 give false terms. and = 3, then More elaborate examples like X 3524 X 3542, X X 3542, X 3524 X , and X 3524 X work equally well, but in the example X 3524 X the multiplicities in some cases are to big. This shows that the requirement in Obs.3.2 for the admissible transitions between consecutive subshapes G p, G p+ is not yet exclusive enough. 4. Observations for the general case This nal section contains some observations, examples and remarks in view of a general rule how to multiply a Schubert polynomial by an X, when D() has more than one component. We begin with a \rule", which works well in many cases, but has also severe defects in other cases to be discussed later on. Observation 4.. (First approximation of a general rule) The following prescriptions give an overall and approximative \rule" for the multiplication of an arbitrary Schubert polynomial by an X in case D() has q > components C ; : : : ; C q : Every component is handled essentially as in Obs.3.2; the relative order of components is such that C i precedes C j, if the leftmost box of C i is left to the leftmost box of C j or, if the 5

16 leftmost boxes of C i and C j are in the same columns, the lower component comes rst. In this way the step tableaux S() is completely determined. For the `Monk step' i from (i?) to (i) suppose that i is in column c(i) and row r(i) of S(). Then i c(i) is given by the requirement that in row r(i) the next gap in D() to the right of column c(i) occurs at column i +. As usual k i i < l i, where k i can be chosen only from the set f; : : : ; i g n F (i) with (4.) F (i) := fk < c(i) j [k; r(i)] 2 D(); and 9h : k < h < c(i); [h; r(i)] =2 D()g being called the -xed set for i. In other words: F (i) is the set of the numbers of those columns left to column c(i), which contain boxes in row r(i) and are separated by a gap from [c(i); r(i)]. Obs.4. already works well in many cases: Example 4.2. We compute X X with = 43625, L() = 30200, S() = , not : , and hence = 2 = 3 =, 4 = 5 = 6 = 4, and F () = F (2) = F (3) = ;, F (4) = F (5) = fg, and F (6) = ;. For the computation it is convenient to draw a box around the numbers on the places F (i). step: : : : j j j j j j j Following simply the \rule" of Obs.4. can result in generating too many permutations, a fact which is partially taken into account by Observation 4.3. Assume that in the situation of Obs.4. i and i+ are in dierent components, and that (i?) (k i ) / (i?) (l i ) for some k i 2 F (i) and arbitrary l i > k i (not necessarily i < l i!). Then the permutation (i?) (k i ; l i ) is generated with a minus sign, i.e. it must be canceled in another place, where it occurs through \regular" generation in accordance to Obs.4.. 6

17 Example 4.4. We compute X 3524 X 432 with = 432, L() = 300, and S() = and step: j3524 3j524 5j j j j j354 3j j j ? 5243 (!) Observation 4.3 marks an important deviation from our general \philosophy" that the (N ) of the product X 0X are reached via admissible paths! But on the other hand it seems very articial and arbitrary to exclude just one of the permutations 5243 in the above example from the nal column. In addition it is unclear, whether only the `xed places' k i 2 F (i) with k i i should be excluded or all k i max F (i). In the rst case there are examples showing that too many permutations are generated and in the second case too less. So one can say that the `interaction' of several components in D() seems to be the cause, which makes it dicult to nd a simple combinatorial rule for the multiplication of Schubert polynomials. Finally the possibility can not yet be ruled out that a combinatorial rule for the computation of a product of Schubert polynomials X 0X has to take into account the properties of both and 0. References [BB] N. Bergeron, S. Billey, RC-Graphs and Schubert polynomials, Experimental Mathematics 2 (993), [C] C. Chevalley, Sur le Decomposition Cellulaires des Espaces G=P, Proceedings of Symposia in Pure Mathematics 56 (994). [Hi] H. Hiller, \Geometry of Coxeter groups", Pitman, Boston, 982. [LS] A. Lascoux, M.-P. Schutzenberger, Decompositions dans l'algebre des dierences divisees, discrete math. 99 (992), [LS2] A. Lascoux, M.-P. Schutzenberger, Polynomes des Schubert, C.R.Acad.Sci.Paris 294 (982),

18 [M] I.G. Macdonald, Schubert polynomials, in: A.D. Kendwell (ed.), \Surveys in Combinatorics", London Math. Soc. Lec. Notes Ser. 66, 73-99, Cambridge University Press, Cambridge 99. [M2] I.G. Macdonald, \Notes on Schubert polynomials", Publications du L.A.C.I.M., vol. 6, Universite du Quebec, Montreal, 99. [M3] I.G. Macdonald, \Symmetric functions and Hall polynomials", Clarendon Press, Oxford 979. [Mo] D. Monk, The geometry of ag manifolds, Proc. London Math. Soc. 3 (959), [Sa] B.E. Sagan, \The Symmetric Group", Wadsworth & Brooks/Cole, Pacic Grove, California (99). [So] F. Sottile, Pieri's formula for ag manifolds and Schubert polynomials, Ann. Inst. Fourier, Grenoble46 (996), [W] R. Winkel, Recursive and combinatorial properties of Schubert polynomials, preprint (995). Institut fur Reine und Angewandte Mathematik, RWTH Aachen, D Aachen, Germany, papers/preprints: winkel/ Current address: Dept. of Mathematics, MIT, 2-390, 77 Mass. Ave., Cambridge, MA , current winkel@math.mit.edu address: winkel@iram.rwth-aachen.de 8