SCHUBERT POLYNOMIALS AND SYMMETRIC FUNCTIONS NOTES FOR THE LISBON COMBINATORICS SUMMER SCHOOL 2012 ALLEN KNUTSON CONTENTS 1. Symmetric polynomials 1 2. Schubert polynomials 3 3. Computing Schubert polynomials 9 4. A positive formula for Schubert polynomials 10 5. Schur polynomials and Schur functions 13 References 16 1. SYMMETRIC POLYNOMIALS In this section R = Z[x1,x2,...,xn] is the ring of polynomials in n variables, with integer coefficients. A polynomial p ∈ R is symmetric if it is unchanged under switching the variables around, i.e. if p(x1,...,xn) = p(xσ(1),xσ(2),...,xσ(n)) for each permutation σ in the sym- metric group Sn. It’s enough to check that p(...,xi,xi+1,...)= p(...,xi+1,xi,...) for each i ∈ [1,n). Some examples: • Any constant function (degree 0 polynomial) is symmetric. • The sum x1 + ... + xn of all the variables is symmetric. k k • The sum x1 + ... + xn of all the kth powers is symmetric. • The sum i≤j xixj of all products, including the squares, is symmetric. • The sum x x of all products, but excluding the squares, is also symmetric. Pi<j i j Let RSn ≤ R denoteP 1 the set of all symmetric polynomials. It is easy to see that it is closed under addition and multiplication, and in particular, is a subring. Motivating question. What polynomials “generate” RSn , analogous to the way that x1,...,xn generate R itself? Polynomial rings such as R have a useful property: every polynomial p can be uniquely written as a sum N pd of homogeneous polynomials pd, in which every monomial in pd has the same degree d. More generally, define a graded ring Q as one that contains a P list (Qd)d∈N of subspaces, such that QdQe ≤ Qd+e and every q ∈ Q is uniquely the sum q = N qd, qd ∈ Qd. DateP: Draft of July 28, 2012. 1More generally, one may write XG to denote the fixed points of an action of a group G on a set X. 1 Z n+k−1 5 3 2 2 Exercise 1.1. Rk has a -basis of size n . (Hint: correspond a monomial like x1x2x4x5, for n = 6, to a word like ∗∗∗∗∗| ∗ ∗ ∗ || ∗ ∗| ∗ ∗||.) Proposition 1.2. The ring RSn of symmetric polynomials is a graded subring, i.e. the homoge- neous pieces of a symmetric polynomial are themselves symmetric. Proof. If p is any polynomial of degree k, we can compute p’s top homogeneous compo- k nent pk as limt p(tx1,tx2,...,txn)/t . If p is symmetric, then the above ratio is symmet- →∞ ric for every t, so is symmetric in the limit. Then p − pk is again symmetric, and if not 0 is of strictly lower degree, so by induction on k its homogeneous pieces are symmetric. Sn Sn Since the degree 1 part (R )1 of R is only 1-dimensional, namely multiples of x1 + x2 + ... + xn, we will need our other generators to be higher degree. To narrow down the problem of giving generators, let’s insist that our generators be homogeneous. Being graded makes it easier for us to look at examples, like n = 2, because we can look one homogeneous degree at a time. Here S2 Z S2 Z S2 2 2 Z (R )0 = · 1, (R )1 = · (x1 + x2), (R )2 = {ax1 + ax2 + bx1x2 : a, b ∈ }. S2 Exercise 1.3. For each polynomial in (R )2, show that it can be written uniquely as a polynomial 2 2 in x1 + x2 and x1 + x2 using rational coefficients. Find one that despite having integer coefficients 2 2 Z itself, cannot be written as p(x1 + x2,x1 + x2) where p(a, b) ∈ [a, b] has integer coefficients. 2 2 So x1 + x2, x1 + x2 will not be a good generating set for us. For each k ≤ n, let e k denote the polynomial ek = xi1 ··· xik , 1≤i <i <...<i ≤n 1 X2 k called the kth elementary symmetric polynomial. Theorem 1.4. Let p be a symmetric polynomial with integer coefficients, i.e. p ∈ RSn . Then p is uniquely expressible as a polynomial in (e1,...,en) with integer coefficients. Put another way, the homomorphism Sn Z[e1,...,en] Z[x1,...,xn] ek 7 xi1 ··· xik → 1≤i <i <...<i ≤n 1 X2 k is an isomorphism of rings. Regarding ek as→ having degree k, it becomes an isomorphism of graded rings. To prove this, we define the lexicographically first monomial m in a nonzero polynomial p, and denote it init p. It is the m with the highest power of x1 available, then among 2 3 ties it has the highest power of x2 available, and so on. (Writing monomials x1x2x3 like x1x1x2x3x3x3, this is almost dictionary order, except that in dictionaries x1x1 comes after x1. It is indeed dictionary order when restricted to monomials of a fixed degree.) k Exercise 1.5. (1) init ek = i=1 xi. (2) init (pq)= init p · init q. Q mi Z (3) If p is symmetric, and init p = c i xi (c ∈ ), then m1 ≥ m2 ≥ ... ≥ mn. Q 2 Proof. It is slightly simpler to break p into the sum of its homogeneous components, and treat each one separately. Which is to say, we reduce to the case that p is homogeneous of some degree k. mi n mi−mi+1 Let init p = c i xi , and E = c i=1 ei (a polynomial, by exercise 1.5 (3)). Then init p = init E, so init p is lex-earlier than init (p − E). Q Q If p−E = 0, we are done. Otherwise it is again homogeneous of degree k. Since there are only finitely many monomials of degree k, and init (p − E) is lex-later than init p, we can use induction to say that p − E is a polynomial in the elementary symmetric polynomials. Hence p itself is also such a polynomial. ni As for uniqueness, observe that each monomial i ei has a different initial term, Pj≤i nj i xi . So we claim that we must use the monomialQ E in writing p, and then invoke induction to get the uniqueness of the expression of p − E. Q n Exercise 1.6. Write i xi as a polynomial in the elementary symmetric polynomials, for n ≤ 4. It is important toP note that it was by no means obvious that RSn would, itself, be iso- morphic to a polynomial ring, as the following exercise shows. Exercise 1.7. Let n = 2, and Z2 = {1,τ} act on R by τ · x = −x, τ · y = −y. Show that the τ-invariant subring RZ2 is generated by x2,xy,y2, and is isomorphic to Z[a, b, c]/hb2 − aci. Show that this graded ring is not graded-isomorphic2 to a polynomial ring. In fact there is a complete classification, due to C. Chevalley, of which finite groups G acting on R and preserving degree have RG isomorphic to a polynomial ring. 2. SCHUBERT POLYNOMIALS One moral of the story so far is that symmetric polynomials have many, many mono- mials in them, and that there are better ways to write down symmetric polynomials than the standard way (namely, by adding up monomials). In this section we’ll consider partially symmetric polynomials, that are symmetric under some exchanges of variables, but not all. Since we won’t need full symmetry, we won’t have to fix a finite number n of variables, and we let R = Z[x1,x2,...] be the polynomial ring in an infinite set of variables. Some references for this theory are [Man01, Mac91, BB93, BJS93, FS94]. 2.1. Divided difference operators. For p ∈ R, and i ∈ N, define the action of the simple reflection operator ri by (ri · p)(x1,...,xi,xi+1,...) := p(x1,...,xi+1,xi,...) so p is symmetric in xi,xi+1 exactly if ri · p = p, or if (1 − ri) · p = 0. It is interesting to note that if q =(1−ri)·p, then ri ·q =−q, and q(x1,...,xi,xi,...)= 0. ′ Using the long division algorithm for polynomials, we can write q = (xi − xi+1)q + c, where deg c< deg(xi − xi+1), i.e. c ∈ Z. Setting xi = xi+1, we learn 0 = 0 + c. 2In fact, there are no isomorphisms at all, even ones that ignore the grading. To show this, one can in some sense force a grading, as follows. Tensor both rings with C, so we can use dimension arguments. 2 C dim 2 3 dimC(m/m ) Then prove that for any maximal ideal m in a polynomial ring over , C(m /m )= 2 , which is part of its being a “regular ring”. But m= ha,b,ci≤ RZ2 doesn’t satisfy this equality. 3 This allows us to define a divided difference operator ∂i on p ∈ R by p − ri · p ∂ip := xi − xi+1 and we see that despite appearances, it is again a polynomial! Exercise 2.1. (1) If p is homogeneous of degree k, then ∂ip is homogeneous of degree k − 1 (or zero). (2) If ∂ip = 0, then ∂i(pq)= p∂iq. (3) More generally (but less usefully) ∂i satisfies the “twisted Leibniz rule”: ∂i(pq)=(∂ip)q +(rip)(∂iq). In these ways, ∂i behaves somewhat like a derivative. (4) Let p be a polynomial such that ∂ip = 0 for all i 6= n. Show that p is a symmetric polynomial in x1,...,xn. 2 Let’s start with p = x1x2 and apply divided differences. This polynomial is symmetric in x3,x4,..