CALTECH ALGEBRAIC GEOMETRY SEMINAR: A GEOMETRIC LITTLEWOOD-RICHARDSON RULE RAVI VAKIL ABSTRACT. I will describe an explicit geometric Littlewood-Richardson rule, interpreted as deforming the intersection of two Schubert varieties so that they break into Schubert varieties. There are no restrictions on the base field, and all multiplicities arising are 1; this is important for applications. This rule should be seen as a generalization of Pieri s rule to arbitrary Schubert classes, by way of explicit homotopies. It has a straightforward bijection to other Littlewood-Richardson rules, such as tableaux and Knutson and Tao s puzzles. This gives the first geometric proof and interpretation of the Littlewood-Richardson rule. It has a host of geometric consequences, which I may describe, time permitting. Buch has suggested an interpretation in K-theory, which has led to a the first conjectural Littlewood- Richardson rule in equivariant K-theory, and a proof of a geometric rule in K-theory (both joint work with Knutson). The rule suggests a natural approach to the open question of finding a Littlewood-Richardson rule for the flag variety, leading to a conjecture, shown to be true up to dimension 5. Finally, the rule suggests approaches to similar open problems, such as Littlewood-Richardson rules for the symplectic Grassmannian and two-step flag varieties. (The original paper appeared in Annals this past year, as did another paper with applications.) CONTENTS 0.1. Pieri s rule 3 1. The rule 3 2. Links to other Littlewood-Richardson rules 4 3. Natural questions 5 4. Generalizations 5 5. Schubert induction 6 5.1. Galois groups of Schubert problems 6 6. Conclusion 6 This is the file caltechag07.tex in my talks directory. The slides are chicago06s and chicago06s2. SLOW DOWN! 20 background 20 rule 20 applications Date: Monday, February 6, 2007. 1
First of all, thanks for having me here to speak. I. Set-up. II. The rule. III. Why you should care: Links to other rules. Applications current and future. I d like to describe to you a geometric version of the Littlewood-Richardson rule. The interpretation of the LR numbers I will use is the geometric one, in terms of the intersection theory of the Grassmannian. In case you are interested in the details, I have a few offprints here. I want going to spend half my time describing the rule, and half my time discussing consequences and further questions. Further geometric consequences are given in a second article. If there is any time left over, which I doubt, I ll say a few words about the proof, which involves the geometry of Bott-Samelson varieties. PAUSE. A Littlewood-Richardson rule is a combinatorial interpretation of Littlewood-Richardson numbers. These numbers have a number of interpretations, in terms of symmetric functions, representation theory, and geometry. I will define these numbers shortly. The first such rule was conjectured by Littlewood-Richardson in the 1930 s. This rule states that Littlewood-Richardson coefficients count certain tableaux. For several reasons, the first proof was not until the 1970 s, by Schutzenberger. Since then, a number of other rules have been proved. One recent one is due to Knutson and Terry Tao, called puzzles. (Sixty-second offer) They have not been geometric in nature, which is also surprising, given that one of the fundamental objects in geometry is the Grassmannian, and one of the fundamental interpretations of Littlewood-Richardson co-efficients is in terms of the cohomology ring of the Grassmannian. Galling to geometers... Given fundamental nature of geometry involved... Here now is the geometric description of LR coefficients. It is in terms of the Grassmannian G(k, n) = G(k 1, n 1), the space of... (e.g. G(1, n) = G(0, n 1) = P n 1 ). To avoid having to answer technical questions, I m going to work over the complex numbers, but this is in fact unnecessary. Works over any field, or indeed any base scheme. Important for applications. Fix a flag in C n, F = {F 0 F 1 F n }. dim F i = i. Consider a k-plane V k C n. How can it intersect the subspaces in F? For example, k = 3, n = 5. Figure: 011223 This corresponds to a k-subset of {1,..., n}. ({1, 3, 5}) This data is usually encoded as a partition. Figure. This is 3 = 2 + 1. It is a subpartition of the k (n k) rectangle. 2
Facts: Ω α (F ) is an affine space C?. Called an open Schubert cell. Codimension is α, so this example is codimension 3. Schubert cycles, i.e. [Ω α ] (no F!), form a basis for H (G). The dual cycle is the complement of the partition in the rectangle. Hence [Ω α ] [Ω β ] = γ cγ αβ [Ω γ] for some c γ αβ, called LR coefficients. They are the structure coefficients for the cohomology ring of the Grassmannian. It is a geometrically straightforward fact that c γ αβ are three general flags, 1 is a non-negative integer: If F, F 2, F 3 c γ αβ = [Ω α] [Ω β ] [Ω γ ] = #Ω α (F 1 ) Ω β(f 2 ) Ω γ (F 3 ). By the Kleiman-Bertini theorem, the right side turns out to be a finite number of points, so the Littlewood-Richardson number is then clearly a non-negative integer. Not obvious with some other descriptions of the rule. 0.1. Pieri s rule. A special case of the Littlewood-Richardson rule is if α = 1 d. In this case, all coefficients are 0 or 1, and there is an easy description that I won t describe now. This version was known earlier, and I believe the first proofs were indeed geometric. Pieri had a proof along the following lines, later Hodge gave a proof in a similar vein; later still, Eisenbud and Harris did too. Consider the subset (indeed subvariety) Ω α (F ) Ω β (F ). Break it into pieces (cleverly), each isomorphic to some Ω γ (F ), and each appearing with multiplicity one. Then [Ω α Ω β ] = c γ αβ [Ω γ]. Strategy discussion: why this shouldn t work in general. What to do to make it work. 20 minutes in BIG PAUSE 1. THE RULE I now want to describe the geometric Littlewood-Richardson rule in this way. I ll give a geometric description. On the slide, I ve displayed a combinatorial description that I won t define, in terms of checkers on a checkerboard. There is a more compact onedimensional notation, but I d like to stick with this one because it is geometrically meaningful. Suppose we have two flags F and M. For convenience, let me take n = 4, and let me draw things projectively, i.e. in 3-space, so we can visualize things. Apology: dimensions. (Figure 1 on slide (blank overlay in red/green).) We are considering Ω α (F ) Ω α (M ) where F and M are transverse. I want to degenerate this intersection by holding F fixed, and moving M: F stands for fixed and M stands for moving. Red means stop and green means go. I ll do it in a series of ( ) n 2 steps, and I ll describe it to you geometrically. First, words: plane; line plane; point line plane. DO IT. 3
Let me describe this degeneration in two other ways. In terms of permutations, this is the standard factorization of the longest word. In terms of computer science, this is a bubble sort of the black checkers. Let s do an example. DO THE KEY EXAMPLE. Let s make this precise. (BLACKBOARD) 1. Varieties. Relative position of the two flags is given by dim F i M j. Relative position of k-plane V k to these two flags is given by dim F i M j V k. We encode this using black and white checkers. To each such data, we have a variety, which is the closure in flag variety cross flag variety cross grassmannian of this locus. 2. Initial position. We started with two k-subsets of {1,..., n}, α and β. This told us how to put white checkers on the board. More precisely, (draw picture). In other words, for two general flags and α and β, Ω α (F ) Ω β (M ) is a two-flag Schubert cell. 3. How to degenerate. Then when we degenerate M in this way, we see that we always get (one or two) more cycles of this form, and the multiplicities are always 1 (theorem). This is true in general. Important fact. 4. Conclude. At the end, the black checkers are lined up on the main diagonal, and the white checkers, in order to be happy, are also on the main diagonal. From this we read off γ. In short, it is a geometric statement, which can be described by a simple rule about pushing black and white checkers around on a board. 20 minutes left. Take stock. Move to slides. 1. Tableaux, puzzles. 2. Question. 3. Applications/Generalizations. 2. LINKS TO OTHER LITTLEWOOD-RICHARDSON RULES So this indeed gives a combinatorial description of the Littlewood-Richardson coefficients. I wish now to link it to other rules. The bijection to tableaux is easy to describe geometrically. Whenever the intersection with the moving flag changes, we put a number in the tableau. Bijection to puzzles (joint with Knutson). There is an equally explicit but less elegant bijection to puzzles. Basically, you play the checkergame, and fill in the puzzle as follows (do it). In particular, a checkergame gives you instructions for filling in a puzzle. 4
Corollaries: (a) Partially filled puzzles have geometric meaning. (b) Tableaux and puzzles are in bijection with solutions to enumerative problems (once branch paths are chosen). 3. NATURAL QUESTIONS Natural questions: Does it scale well? Yes. Triality? Yes. Is c γ αβ PAUSE. = cγ βα? Not clear. 4. GENERALIZATIONS One motivation for the Geometric Littlewood-Richardson rule is that it should generalize well to other important geometric situations. I ll now briefly describe some potential applications. K-theory. A. Buch proved the first LR rule in K-theory, involving set-valued tablueax. There is also a checker K-theoretic LR rule as well, which was conjectured by Buch. You can prove it by giving a bijection to his set-valued tableaux. Thanks to the bijection to puzzles, we get a new puzzle-result: there is a new K-theory puzzle piece (describe it). As a result, we get triality of K-theory LR co-efficients, which was initially surprising to the people working in the area. But once Anders saw the statement, he was able to give a fast proof. There is a stronger statement buried in Buch s statement, which is as yet conjectural. When the cycle breaks into two pieces, the scheme-theoretic intersection is the set-theoretic intersection. Move to slides. Equivariant K-theory. With A. Knutson, in progress. Doing this on the flag variety. Important open question: find a LR rule for the flag manifold. (Multiply Schubert polynomials together.) This method looks promising, but I ve been unable to complete it, after a great deal of work. Inexplicit Conjecture. The corresponding important fact is true. True for n 5. Unclear for n = 6. This is on one hand serious evidence, but on the other hand you should still be quite suspicious. The combinatorial objects that arise here can be expressed in terms of checkers, or in terms of wiring diagrams that bear a striking resemblance to rc-graphs, although I can t make out if there is actually a connection. Other groups. These methods may apply to other groups where Littlewood-Richardson rules are not known. For example, for the symplectic Grassmannian, there are only rules in the Lagrangian and Pieri cases. L. Mihalcea has made progress in finding a Geometric 5
Littlewood-Richardson rule in the Lagrangian case, and has suggested that a similar algorithm should exist in general. His ideas are very interesting, and I d like to advertise them here. Quantum cohomology. With Ciocan-Fontanine. Alternatively, the quantum cohomology of the Grassmannian can be translated into classical questions in the enumerative geometry of surfaces. One may hope that degeneration methods will apply. This perspective is being pursued (with different motivation) independently by I. Coskun (for rational scrolls), D. Avritzer, and M. Honsen (on Veronese surfaces). SKIP: (e) Two-step partial flag manifold and similar homogeneous spaces. (f) Polygonalization of Eisenbud and Harris path? Scherbak. Implies Shapiro-Shapiro asymptotically. [(g) Fomin-Fulton-Li-Poon.] (g) Intersections of Schubert varieties in GL n /B and other permutation array varieties (with S. Billey, in preparation). END SKIP. 5. SCHUBERT INDUCTION Reals. (One set of flags.) Kleiman-Bertini theorem. Positive characteristic. Finite fields. 5.1. Galois groups of Schubert problems. Serre. Three definitions: geo, alg, arithm. Harm. Question 1: In P 7, fix four P 3 s V 1,..., V 4. How many P 3 s meet V 1, V 2, V 3, V 4 in lines. Answer: 6. Galois group = S 4, edges of a tetrahedron. Why? Question 2: c (6,6,4,4,2,2) (4,4,2,2),(4,4,2,2) = 6. What does this mean? 6. CONCLUSION I hope I ve given you some idea about this geo LR rule, which is reasonably simple to describe, and the central idea of degenerating in this particular way, getting multiplicity 1 pieces, and some of the consequences that arise. Thank you for your patience. 6