Grassmannians: The First Example of a Moduli Space

GRASSMANNIANS: THE FIRST EXAMPLE OF A MODULI SPACE 1. What is this course about? Many objects of algebraic geometry, such as subspaces of a linear space, smooth curves of genus g, or stable vector bundles on a curve, themselves vary in alge­ braically defined families. Moduli theory studies such families of algebraic objects. Roughly speaking a moduli problem is the problem of understanding a given geometrically meaningful functor from the category of schemes to sets. To make this more concrete consider the following three functors. Example 1.1 (Example 1: The Grassmannian Functor.). Let S be a scheme, E a vector bundle on S and k a positive integer less than the rank of E. Let Gr(k, S, E) : {Schemes/S} � {sets} be the contravariant functor that associates to an S-scheme X subvector bundles of rank k of X ×S E. Example 1.2 (Example 2: The Hilbert Functor.). Let X � S be a projective scheme, O(1) a relatively ample line bundle and P a fixed polynomial. Let HilbP (X/S) : {Schemes/S} � {sets} be the contravariant functor that associates to an S scheme Y the subschemes of X ×S Y which are proper and flat over Y and have the Hilbert polynomial P . Example 1.3 (Example 3: Moduli of stable curves.). Let Mg : {Schemes} � {sets} be the functor that assigns to a scheme Z the set of families (up to isomorphism) X � Z flat over Z whose fibers are stable curves of genus g. Each of the functors in the three examples above poses a moduli problem. The first step in the solution of such a problem is to construct a smooth, projective variety/proper scheme/ proper Deligne-Mumford stack that represents the functor finely/coarsely. Definition 1.4. Given a contravariant functor F from schemes over S to sets, we say that a scheme X(F ) over S and an element U (F ) ∩ F (X(F )) represents the functor finely if for every S scheme Y the map HomS (Y, X(F )) � F (Y ) given by g � g � U (F ) is an isomorphism. 1 The best answer one can usually hope for (such as in Examples 1 and 2) is that there is a scheme (hopefully proper) representing the functor. There may not be such a scheme. For instance for the functor in Example 3 there does not exist a fine moduli scheme representing the functor. In such cases we represent the functor either in a different category or we relax the conditions that we impose on the representing scheme. The most common alternatives are to work with stacks or to ask for the moduli space to only coarsely represent the functor. Definition 1.5. Given a contravariant functor F from schemes over S to sets, we say that a scheme X(F ) over S coarsely represents the functor F if there is a natural transformation of functors � : F � HomS (�, X(F )) such that (1) �(spec(k)) : F (spec(k)) � HomS (spec(k), X(F )) is a bijection for every algebraically closed field k, (2) For any S-scheme Y and any natural transformation Σ : F � HomS (�, Y ), there is a unique natural transformation Ω : HomS (�, X(F )) � HomS (�, Y ) such that Σ = Ω ∧ �. Finding a representing scheme/stack, a moduli space, is only the first step of a moduli problem. Usually the motivation for constructing a moduli space is to understand the objects this space parameterizes. This in turn requires a good knowledge of the geometry of the moduli space. Among the questions that arise about these moduli spaces are: (1) Is the moduli space proper? If not, does it have a modular compactification? Is the moduli space projective? (2) What is the dimension of the moduli space? Is it connected? Is it irre­ ducible? What are its singularities? (3) What is the cohomology/Chow ring of the moduli space? (4) What is the Picard group of the moduli space? Assuming the moduli space is projective, which of the divisors are ample? Which of the divisors are effective? (5) Can the moduli space be rationally parameterized? What is the Kodaira dimension of the moduli space? The second step of the moduli problem is answering as many of these questions as possible. The focus of this course will be the second step of the moduli problem. In this course we will not concentrate on the constructions of the moduli spaces. We will often stop at outlining the main steps of the constructions only in so far as they help us understand the geometry. We will spend most of the time talking about the explicit geometry of these moduli spaces. We begin our study with the Grassmannian. The Grassmannian is the scheme that represents the functor in Example 1. Grassmannians lie at the heart of moduli theory. Their existence is the first step for the proof of the existence of the Hilbert scheme. Many moduli spaces in turn can be constructed using the Hilbert scheme. On the other hand, the Grassmannians are sufficiently simple that their geometry is well-understood. Many of the constructions for understanding the geometry of other moduli spaces, such as the moduli space of stable curves, imitates the techniques used in the case of Grassmannians. This motivates us to begin our exploration with the Grassmannian. 2 Additional references: For a more detailed introduction to moduli problems you might want to read [HM] Chapter 1 Section A, [H] Lecture 21, [EH] Section VI and [K] Section I.1. 2. Preliminaries about the Grassmannian Good references for this section (in random order) are [H] Lectures 6 and 16, [GH] I.5 and [Ful2] Chapter 14, [Kl2] and [KL]. Let G(k, n) denote the classical Grassmannian that parameterizes k-dimensional linear subspaces of a fixed n-dimensional vector space V . G(k, n) naturally carries the structure of a smooth, projective variety. It is often convenient to think of G(k, n) as the parameter space of k−1-dimensional projective linear spaces in P n−1 . When we use this point of view, we will denote the Grassmannian by G(k −1, n−1). It is easy to give G(k, n) the structure of an abstract variety. In case V = C n , G(k, n) becomes a complex manifold under this structure. Given a k-dimensional subspace � of V , we can represent it by a k × n matrix. Choose a basis for � and write them as the row vectors of the matrix. GL(k) acts on the left by multipli­ cation. Two k × n matrices represent the same linear space if and only if they are related by this action of GL(k). Since the k vectors span �, in the matrix repre­ sentation there must exist a non-vanishing k × k minor. Suppose we look at those matrices that have a fixed non-vanishing k × k minor. We can normalize that this submatrix is the identity matrix. This gives a unique representation for �. In this representation the remaining entries are free to vary. The space of such matrices is isomorphic to Ak(n−k) . In case V = Cn the transition functions are clearly holo­ morphic. We thus obtain the structure of a complex manifold of dimension k(n−k) on the Grassmannian G(k, n). The Grassmannian is compact and connected (for example, the unitary group U (n) maps continuously onto G(k, n)). The cohomology/Chow ring of the Grassmannian can be very explicitly de­ scribed. Fix a flag F• : 0 = F 0 ≥ F 1 ≥ · · · ≥ F n = V in the vector space V . Recall that a flag is a nested sequence of vector subspaces of V where the difference in dimension of two consecutive vector spaces is one. Given a partition � with k parts satisfying the conditions n − k → �1 → �2 → · · · → �k → 0, we can define a subvariety of the Grassmannian called the Schubert variety ��1 ,...,�k (F• ) of type � with respect to the flag F• to be the closure of �0�1 ,...,�k (F• ) := { [�] ∩ G(k, n) : dim(� ∗ Fn−k+i−�i ) = i }. This closure is obtained by turning all = to → in the rank conditions. The homology class of a Schubert variety does not depend on the choice of flag. For each partition satisfying the above properties, we get a homology class. A word of caution: Most Schubert varieties in a Grassmannian are singular. Fix the standard flag for Cn where Fi =< e1 , . . . , ei >. Let � be a k-plane in the open part of the Schubert variety ��1 ,...,�k (F• ) defined with respect to this flag. We can normalize � ∗ Fn−k+i−�i so that < vi , en−k+j−�j >= 0 for j < i and < vi , en−k+i−�i >= 1. Thus we get a unique matrix representation for � and 3 see that �0�1 ,...,�k (F• ) � = Ak(n−k)− i �i . In other words, Schubert varieties give a cell-decomposition of G(k, n) with only even dimensional cells. It follows that the classes of Schubert varieties generates the homology of G(k, n). Applying Poincaré duality we obtain the following fundamental theorem about the cohomology of G(k, n). We will denote the cohomology class that corresponds to the Schubert variety ��1 ,...,�k by ��1 ,...,�k . We often omit the indices that are zero. P Theorem 2.1. The Poincar´e duals of the classes of Schubert varieties give an additive basis of the cohomology of the Grassmannian. Example 2.2. Let us consider the case G(2, 4) = G(1, 3). This variety geometri­ cally corresponds to the variety of lines in P3 . The Schubert varieties are in this case given by �1 in codimension 1, �1,1 and �2 in codimension 2, �2,1 in codi­ mension 3 and �2,2 in codimension 4. Of course, all the codimensions are complex codimensions. A flag in P3 corresponds to a choice of point q contained in a line l contained in a plane P contained in P3 . �1 parameterizes lines that intersect l. �2 parameterizes lines that contain q. �1,1 parameterizes lines that are contained in P . �2,1 parameterizes lines that are contained in P and contain q. Since the cohomology of Grassmannians is generated by Schubert cycles, given two Schubert cycles �� and �µ , their product in the cohomology ring can be ex­ pressed as a linear combination of Schubert cycles. � � �� · � µ = c�,µ �� � � The structure constants c�,µ of the cohomology ring with respect to the Schubert basis are known as Littlewood - Richardson coefficients. Example 2.2 continued. Let us work out the Littlewood - Richardson coefficients of G(2, 4) = G(1, 3). All but one of the calculations are easy. It is simplest to work dually with the intersection of Schubert varieties. Suppose we wanted to calculate �2 ∗ �2 . �2 is the class of lines that pass through a point. If we take points, there will be a unique line containing them both. We conclude that �2 ∗ �2 = �2,2 . Similarly �1,1 ∗ �1,1 = �2,2 , because there is a unique line contained in two distinct planes in P3 . On the other hand �1,1 ∗�2 = 0 since there will not be a line contained in a plane and passing through a point not contained in the plane. The hardest class to compute is �1 ∗ �1 . We know that the class is expressible as a linear combination of �1,1 and �2 . We just saw that both these cycles are selfdual. In order to compute the coefficient we can calculate the triple intersection. �1 ∗ �1 ∗ �2 is the set of lines that meet two lines l1 , l2 and contain a point q. There is a unique such line given by ql1 ∗ ql2 . The other coefficient can be similarly computed to see �12 = �1,1 + �2 . Exercise 2.3. Work out the multiplicative structure of the cohomology ring of G(2, 4) = G(1, 3), G(2, 5) = G(1, 4) and G(3, 6) = G(2, 5). Exercise 2.4. Show that the dual of the Schubert cycle ��1 ,...,�k is the Schubert cycle �n−k−�k ,...,n−k−�1 . Conclude that the Littlewood - Richardson coefficient c��,µ may be computed as the triple product �� · �µ · �� � . The method of undetermined coefficients we just employed is a powerful tech­ nique for calculating the classes of subvarieties of the Grassmannian. Let us do an example to show another use of the technique. 4 Example 2.5. How many lines are contained in the intersection of two general quadric hypersurfaces in P4 ? In order to work out this problem we can calculate the class of lines contained in a quadric hypersurface in P4 and square the class. The dimension of the space of lines on a quadric hypersurface is 3. The classes of dimension 3 in G(1, 4) are given by �3 and �2,1 . We can, therefore, write this class as a�3 + b�2,1 . The coefficient of �3 is zero because �3 is self-dual and corresponds to lines that pass through a point. As long as the quadric hypersurface does not contain the point, the intersection will be zero. On the other hand, b = 4. �2,1 parameterizes lines in P4 that intersect a P1 and are contained in a P3 containing the P1 . The intersection of the quadric hypersurface with the P3 is a quadric surface. The lines have to be contained in this surface and must pass through the two points of intersection of the P1 with the quadric surface. There are four such lines. We conclude that there are 16 lines that are contained in the intersection of two general quadric hypersurfaces in P4 . Another way to verify this fact is to observe that such an intersection is a quartic Del Pezzo surface. Such a surface is the blow-up of P2 at 5 general points embedded by its anti-canonical linear system. The lines in this embedding correspond to the (−1)-curves on the surface. It is well-known that the number of (−1)-curves on this surface is 16 (see for example [Ha] Chapter 5). There is one issue that requires some attention. So far we have pretended that all the intersections are transverse. This is indeed the case. We can either explicitly calculate the tangent spaces to check that the intersection is transverse or we can appeal to a general theorem that guarantees the result. Since the theorem is very useful, we reproduce its statement here. However, be warned that the theorem in the form stated holds only in characteristic zero. For a proof see [Kl1] or [Ha] Theorem III.10.8. Theorem 2.6. (Kleiman) Assume we are working over an algebraically closed field of characteristic zero. Let G be an integral algebraic group scheme, X an integral algebraic scheme with a transitive G action. Let f : Y � X and g : Z � X be two maps of integral algebraic schemes. For each rational element of g ∩ G, denote by gY the X-scheme given by y ∈� gf (y). Then there exists a dense open subset U of G such that for every rational element g ∩ U , the fiber product (gY ) × X Z is either empty or equidimensional of the expected dimension dim(Y ) + dim(Z) − dim(X). Furthermore, if Y and Z are regular, for a dense open set this fibered product is regular. Proof. The theorem follows from the following lemma. Lemma 2.7. Suppose all the schemes in the following diagram are integral over an algebraically closed field of characteristic zero. W �� � � �� �� � Z p S �� �� � ��� r q X 5 If q is flat, then there exists a dense open subset of S such that p−1 (s) ×X Z is empty or equidimensional of dimension dim(p−1 (s)) + dim(Z) − dim(X). If in addition, Z is regular and q has regular fibers, then p−1 (s) ×X Z is regular for a dense open subset of S. The theorem follows by taking S = G, W = G × Y and q : G × Y � X given by q(g, y) = gf (y). The lemma follows by flatness and generic smoothness. More precisely, since q is flat, the fibers of q are equidimensional of dimension dim(W ) − dim(X). By base change the induced map W ×X Z � Z is also flat, hence the fibers have dimension dim(W ×X Z) − dim(Z). Consequently, dim(W ×X Z) = dim(W ) + dim(Z) − dim(X). There is an open subset U1 ≥ S over which p is flat, so the fibers are either empty or equidimensional with dimension dim(W ) − dim(S). Similarly there is an open subset U2 ≥ S, where the fibers of p ∧ prW : X ×X Z � S is either empty or equidimensional of dimension dim(X ×X Z) − dim(S). The first part of the lemma follows by taking U = U1 ∗ U2 and combining these dimension statements. The second statement follows by generic smoothness. This is where we use the assumption that the characteristic is zero. � The Grassmannians G(k, n) are homogeneous under the action of GL(n). Hence Kleiman’s Theorem easily implies the transversality of intersections in many cases. We now give two presentations for the cohomology ring of the Grassmannian. These presentations are useful for theoretical computations. However, we will soon develop Littlewood - Richardson rules, positive combinatorial rules for computing Littlewood - Richardson coefficients, that are much more effective in computing and understanding the structure of the cohomology ring of G(k, n). One extremely useful way comes from considering the universal exact sequence of bundles on G(k, n). Let T denote the tautological bundle over G(k, n). Recall that the fiber of T over a point [�] is the vector subspace � of V . There is a natural inclusion 0�T �V �Q�0 with quotient bundle Q. Theorem 2.8. As a ring the cohomology ring of G(k, n) is isomorphic to R[c1 (T ), . . . , ck (T ), c1 (Q), . . . , cn−k (Q)]/(c(T )c(Q) = 1). Moreover, the chern classes of the Quotient bundle generate the cohomology ring. The chern classes of the tautological bundle and the quotient bundle are easy to see in terms of Schubert cycles. As an exercise prove the following proposition: Proposition 2.9. The chern classes of the tautological bundle are given as follows: ci (T ) = (−1)i �1,...,1 where there are i ones. The chern classes of the quotient bundle are given by ci (Q) = �i . 6 The Schubert cycles �i where all the parts of the partition except for the first are zero are called special Schubert cycles. It is easy to calculate the product of special Schubert cycles. Pieri’s rule gives an algorithm for computing these products. In fact, Pieri’s rule gives an algorithm for computing the product of any Schubert cycle with a special Schubert cycle. Theorem 2.10 (Pieri’s formula). Let �� be a special Schubert cycle. Suppose �µ is any Schubert cycle with parts µ1 , . . . , µk . Then � (1) �� · � µ = �� µi ��i �µP i−1 P �i =�+ µi The special Schubert cycles generate the cohomology ring of the Grassmannian. In order to prove this we have to express every Schubert cycle ��1 ,...,�k as a linear combination of products of special Schubert cycles. Exercise 2.11. Using Pieri’s formula prove the following identity (−1)k ��1 ,...,�k = k � (−1)j ��1 ,...,�j−1 ,�j+1 −1,...,�k −1 · ��j +k−j j=1 Using this relation and induction obtain the following formula for the class of any Schubert cycle in terms of special Schubert cycles. Theorem 2.12 (Giambelli’s formula). Any Schubert cycle may be expressed as a linear combination of products of special Schubert cycles as follows � � � ��1 +1 ��1 +2 . . . ��1 +k−1 �� � � �1 � � � �2 ��2 +1 . . . ��2 +k−2 �� ��1 ,...,�k = �� �2 −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .�� � ���k −k+1 ��k −k+2 ��k −k+3 . . . � �k � Exercise 2.13. Use Giambelli’s formula to express �3,2,1 in G(4, 8) in terms of special Schubert cycles. Using Pieri’s rule find the class of its square. Pieri’s formula and Giambelli’s formula together give an algorithm for comput­ ing the cup product of any two Schubert cycles. Unfortunately, in practice this algorithm is hard to implement. We will rectify this problem shortly. So far we have treated the Grassmannian simply as a complex manifold. For the sake of completeness, we recall how to endow it with the structure of a smooth, pro� jective variety. Using the Plücker coordinates we can embed G(k, V ) into P( k V ). Given a k-plane � we can choose a basis for it v1 , . . . , vk . Then we can define the �k map P l : G(k, n) � P( V ) by sending the k-plane � to v1 ≤ · · · ≤ vk . A change of basis changes the image by the determinant of the matrix giving the change of �k basis. Hence the map is well-defined as a point of P( V ). The map is injective since we can recover � from its image p = [v1 ≤ · · · ≤ �k vk ] ∩ P( V ) as the set of all vectors v ∩ V such that v ≤ v1 ≤ · · · ≤ vk = 0. �k A point of P( V ) is in the image of this map if and only if the representative pi1 ,...,ik e1 ≤ · · · ≤ eik is completely decomposable. It is not hard to characterize �k the subvariety of P( V ) corresponding to completely decomposable elements. An 7 �k element x ∩ V is completely decomposable if and only if < u, x > ≤x = 0 for �k−1 � every u ∩ V . Writing this in coordinates we obtain the Plücker relations k+1 � (−1)s pi1 ,...,ir−1 ,jt pj1 ,...,jˆt ,...jr+1 = 0. s=1 These Plücker relations generate the ideal of the Grassmannian. Everyone’s favorite example is G(2, 4). In that case there is a unique Plücker relation p12 p34 − p13 p24 + p14 p23 = 0. Hence the Plücker map embeds G(2, 4) in P5 as a smooth quadric hypersurface. Exercise 2.14. Show that the locus where a Plücker coordinate vanishes corre­ sponds to a Schubert variety �1 . Observe that the class of �1 generates the second homology of the Grassmannian. In particular, the Picard group is isomorphic to Z. Conclude that OG(k,n) (�1 ) is the very ample generator of the Picard group and it gives rise to the Plücker embedding. We can compute the degree of the Grassmannian G(k, n) under the Plücker k(n−k) . When k = 2, this computation is embedding. The answer is provided by �1 relatively easy to carry out. By Pieri’s formula �1 times any cycle in G(2, n) either increases the first index of the cycle or it increases the second index provided that it is less than the first index. This means that the degree of the Grassmannian G(2, n) is the number of ways of walking from one corner of an (n − 2) × (n − 2) to the opposite corner without crossing the diagonal. This is well-known to be the Catalan number (2(n − 2))! . (n − 2)!(n − 1)! The general formula is more involved. The degree of G(k, n) is given by (k(n − k))! k � i=1 (i − 1)! . (n − k + i − 1)! The local structure of the Grassmannian. The tangent bundle of the Grass­ mannian has a simple intrinsic description in terms of the tautological bundle T and the quotient bundle Q. There is a natural identification of the tangent bundle of the Grassmannian with homomorphisms from T to Q, in other words T G(k, n) = Hom(T, Q). In particular, the tangent space to the Grassmannian at a point [�] is given by Hom(�, V /�). One way to realize this identification is to note that the Grassman­ nian is a homogeneous space for GL(n). The tangent space at a point may be naturally identified with quotient of the Lie algebra of GL(n) by the Lie algebra of the stabilizer. The Lie algebra of GL(n) is the endomorphisms of V . Those that stabilize � are those homomorphisms � : V � V such that �(�) ≥ �. These homomorphisms are precisely homomorphisms Hom(�, V /�). Exercise 2.15. Use the above description to obtain a description of the tangent space to the Schubert variety ��1 ,...,�k at a smooth point [�] of the variety. 8 We can use the description of the tangent space to check that the intersection of Schubert cycles in previous calculations were indeed transverse. For example, suppose we take the intersection of two Schubert varieties �1 in G(1, 3) defined with respect to two skew-lines. Then the intersection is a smooth variety. In vector space notation, we can assume that the conditions are imposed by two non-intersecting two-dimensional vector spaces V1 and V2 . Suppose a 2-dimensional vector space � meets each in dimension 1. The tangent space to � at the intersection is given by � ∩ Hom(�, V /�) such that �(� ∗ Vi ) ≥ [Vi ] ∩ V /�. As long as V1 and V2 do not intersect, � has exactly a one-dimensional intersection with each of Vi and these span �. On the other hand, the quotient of Vi in V /� is one-dimensional. We conclude that the dimension of such homomorphisms is 2. Since this is equal to the dimension of the variety, we deduce that the variety is smooth. Exercise 2.16. Carry out a similar analysis for the other examples we did above. Definition 2.17. Let S be a scheme, E a vector bundle on S and k a natural number less than or equal to the rank of E. The functor Gr(k, E) : {schemes overS} � {sets} associates to every S scheme X the set of rank k subvector bundles of E × S X. Theorem 2.18. The functor Gr(k, E) is represented by a scheme GS (k, E) and a subvector bundle U ≥ E ×S GS (k, E) of rank k. 3. A Littlewood-Richardson rule Positive combinatorial rules for determining Littlewood - Richardson coefficients are known as Littlewood - Richardson rules. As an introduction to the degeneration techniques that we will employ through out this course, we give a Littlewood ­ Richardson rule for the Grassmannian. There are many Littlewood - Richardson rules for the Grassmannian. You can find other Littlewood - Richardson rules in [Ful1], [V1], [KT]. The rule we will develop here is a geometric Littlewood - Richardson rule. These rules have many applications in geometry. For some examples of applications to positive character­ istic, Schubert calculus over R and monodromy groups see [V2]. The fundamental example. Consider calculating �12 in G(2, 4). Geometrically we would like to calculate the class of two dimensional linear spaces that meet two general two dimensional linear spaces in a four dimensional vector space. Projec­ tivizing this question is equivalent to asking for the class of lines in P3 intersecting two general lines. The idea underlying the approach to answering this question is classical. While it is hard to see the Schubert cycles that constitute this intersection when the two lines that define the two Schubert cycles are general, the result becomes easier if the lines are in special position. To put the lines l1 and l2 in a special position fix a plane containing l1 and rotate it about a point on it, so that it intersects l2 . As long as l1 and l2 do not intersect, they are in general position since the automorphism group of P3 acts transitively 9 on pairs of skew lines. However, when l1 and l2 intersect, then they are no longer in general position. We can ask the following fundamental question: What is the limiting position of the lines that intersect both l1 and l2 ? Since intersecting the lines are closed conditions, any limit line has to continue to intersect l1 and l2 . There are two ways that a line can intersect two intersecting lines in P3 . Either the line passes through their intersection point, or if it does not pass through its intersection point then it must lie in the plane spanned by the two lines. Note that these are both Schubert cycles. Since their dimensions are equal to the dimension of the original variety, the class of the original variety has to be the sum of multiples of these two Schubert cycles. We can determine that the multiplicities are one as follows. The tangent space to the Grassmannian G(k, n) at a point Φ is given by It suffices then to check that the two cycles intersect transversely at a general point of each of the Schubert cycles. A Mondrian tableau associated to a Schubert class ��1 ,··· ,�k in G(k, n) is a col­ lection of k nested squares labeled by integers 1, . . . , k where the j-th box has size n − k + j − �j and a box of smaller index is contained in every box of larger index. Figure 1 depicts a Mondrian tableau for �2,1 in G(3, 6). A3 A2 unit size A1 Figure 1. The Mondrian tableau associated to �2,1 in G(3, 6). In Mondrian tableaux a box of side length s denotes a vector space of dimension s. If a box S1 is contained in another box S2 , then the linear space represented by S1 is a subspace of the linear space represented by S2 . The reader should think of unit squares along the anti-diagonal as giving a basis of the underlying vector space. The vector space represented by a box is the span of the basis elements it contains. In a Mondrian tableau associated to �� the k-plane is required to meet the vector space represented by a box in dimension equal to the number of boxes contained in that box (including itself). We will denote boxes in a Mondrian tableau by capital letters in the math font (e.g. Ai ) and the vector spaces they represent by the corresponding letter in Roman font (e.g. Ai ). We stress that any nested sequence of a boxes that have their centers along the anti-diagonal defines a Schubert cycle. The boxes need not be left or right aligned. The game. To multiply two Schubert classes �� and �µ in G(k, n) we place the tableau associated to � starting from the lower left hand corner and the tableau associated to µ starting from the upper right hand corner of an n × n square. The squares in the � (µ) tableau are all left (respectively, right) aligned with respect to the n × n square. We will denote the boxes corresponding to � and µ by Ai and Bj , respectively. The left panel in Figure 2 shows the initial tableau for the multiplication �2,1,1 · �1,1,1 in G(3, 6). 10 B3 B2 B1 A3 A2 A1 Outer box B3 B2 B1 A3 A2 A1 Unit size Figure 2. An application of the OB rule. Initially the two Schubert cycles are defined with respect to two transverse flags. If the intersection of the two Schubert cycles is non-empty, then the Schubert cycles have to satisfy certain conditions. A preliminary rule (MM rule) guarantees that these conditions are satisfied. Then there are some simplifications that reduce the problem to a smaller problem. The OB and S rules give these simplifications. • The MM rule. We check that Ai intersects Bk−i+1 in a square of side length at least one for every i between 1 and k. If not, we stop. The Schubert cycles have empty intersection. In other words, the class of the intersection is zero. In a k-dimensional vector space V k every i-dimensional subspace (such as V k ∗ Ai ) Must Meet every k − i + 1-dimensional subspace (such as V k ∗ Bk−i+1 ) in at least a line. The intersection of two Schubert cycles is zero if and only if the initial tableau formed by the two cycles does not satisfy the MM rule. • The OB rule. We call the intersection of Ak and Bk the Outer Box of the tableau. We replace every square with its intersection with the outer box. Since the k-planes are contained in both Ak and Bk , they must be contained in their intersection. Figure 2 shows an example in G(3, 6). • The S rule. We check that Ai and Ba−i touch or have a common square. If not, we remove the rows and columns between these squares as shown in Figure 3. This rule corresponds to the fact that an a-dimensional vector space lies in the Span of any two of its subspaces of complementary dimension whose only intersection is the origin. This rule removes any basis element of V that is not needed in expressing the a-planes parameterized by the intersection of the two Schubert varieties. B1 A2 Figure 3. Adjusting the span of the linear constraints. Once we have performed these preliminary steps, we will inductively build a new flag (the D flag) by degenerating the two flags (the A and B flags). At each stage 11 of the game we will have a partially built new flag (depicted by D boxes that arise as intersections of A and B boxes) and partially remaining A and B flags (depicted by boxes Ai , . . . , Aa and Bk , Bk−i , . . . , B1 ). After nesting the D boxes, we will increase the dimension of the intersection of Ai with Bk−i by one in order of increasing i. We will depict this move in the Mondrian tableau by sliding Ai anti-diagonally up by one unit. Assuming that there are no boxes left justified with Ai , the corresponding degeneration can be described as follows: Let s be the side-length of Ai and suppose that initially Ai and Ba−i intersect in a square of side-length r. There is a family of s-dimensional linear spaces Ai (t) parameterized by an open subset 0 ∩ U ≥ P1 such that over the points t ∩ U with t ⊂= 0, the dimension of intersection Ai (t) ∗ Bk−i is equal to r and when t = 0, the dimension of intersection Ai (0) ∗ Ba−i is r + 1. Denoting the basis vectors represented by the unit squares along the diagonal by e1 , . . . , en , we explicitly take the family to be Ai (t) = the span of {(te1 + (1 − t)es+1 , e2 , . . . , es }. When t = 1, we have our original vector space Ai represented by the old position of the box Ai . When t = 0, we have the new vector space Ai (0) represented by the new position of the box Ai . When t = 0, the intersection of Schubert varieties defined with respect to the A and B flags either remains irreducible or breaks into two irreducible components. The LR rule records these possibilities and can be informally phrased as: If the a-planes in the limit do not intersect Ai (0) ∗ Bk−i , then they must be contained in their new span. The main work in establishing the rule rests in describing which varieties (equiva­ lently which Mondrian tableaux) occur as a result of the degenerations (equivalently moves). Very generally we can define a Mondrian tableau in G(k, n) to be a collec­ tion of k boxes contained in an n × n box satisfying the following two properties: (1) None of the boxes are equal to the span of the boxes contained in them. (2) Let S1 and S2 be any two boxes in the tableau. If the number of boxes contained in their span but not contained in S1 is r, then the side length of S1 is at least r less than the side length of their span. We can associate an irreducible subvariety of the Grassmannian G(k, n) to such a tableau. We first define an open subset of the variety by requiring the k-planes to meet the vector spaces represented by each box in dimension equal to the number of boxes contained in that box (including itself). We further require the vector sub­ spaces of the k-planes contained in the vector spaces represented by any two boxes to only meet along the subspaces contained in subspaces contained in boxes com­ mon to both of the boxes and otherwise to be independent. The variety associated to the generalized Mondrian tableau is the closure in G(k, n) of the quasi-projective variety parameterizing such k-planes. The intersection of two Schubert varieties can be turned to such a tableau by replacing the boxes Ai and Bj by the boxes consisting of the intersections Ai ∗ Bk−i+1 . Here we will not discuss the rule that expresses the classes of the varieties defined by these very general tableaux as a sum of Schubert varieties. When we resolve the intersection of Schubert varieties into a union of Schubert varieties, only very few of these varieties occur. During the game the Mondrian tableaux 12 that occur have more structure. The admissible tableaux characterize the ones that occur. A Mondrian tableau is admissible for G(k, n) if the squares that constitute the tableau (except for the outer box) are uniquely labeled as an indexed A, B or D box such that (1) The boxes Ak = Bk form the outer box. They have side length m ∼ n and contain the entire tableau. (2) The A boxes are all nested, distinct, left aligned and strictly contain all the D boxes. If the number of D boxes is i − 1 < k, then the A boxes are Ai , Ai+1 , . . . , Ak with the smaller index corresponding to the smaller box. (In particular, the number of A boxes is k − i + 1, hence the number of A and D boxes add up to k.) (3) The B boxes are all nested, distinct and right aligned. They are labeled Bk , Bk−i , Bk−i−1 , . . . , B1 , where a smaller box has the smaller index. (In particular, the number of B boxes equals the number of A boxes.) The A and B boxes satisfy the MM and S rules. The D boxes may intersect Bk−i , but none are contained in Bk−i . The side length of Bk−i is at least i units smaller than the side length of the outer box and at least h units smaller than the side length of the box spanned by the boxes Ds for 1 ∼ s ∼ h and Bk−i for every 1 ∼ h ∼ i − 1. (4) The D boxes are labeled D1 , . . . , Di−1 . They do not need to be nested; however, there can be at most one unnested D box. An unnested D box is defined to be a D box that does not contain every D box of smaller index. More precisely, if Dj does not contain all the D boxes of smaller index, then it does not contain any of the D boxes of smaller index; it is contained in every D box of larger index; and Dh ≥ Dk for every h < k as long as h and k are different from j. All the D boxes of index lower than j are to the lower left of Dj . Dj−1 and Dj share a common square or corner. If there is an unnested D box Dj , the D or A box with one larger index is at least one larger than the span of the D boxes contained in it. The side length of Dj is at least i smaller than the side length of the square spanned by Di and Dj for every i < j. Given an admissible Mondrian tableau there is a corresponding subvariety of G(k, n). The corresponding subvariety is defined as the closure of the locus of kplanes that satisfy certain numerical conditions with respect to the vector spaces represented by the A, B and D boxes. Precisely, the variety associated to an admissible Mondrian tableau is the closure of the locus of k-planes that intersect the vector spaces represented by the boxes Ds , s = 1, . . . , i − 1, At , t = i, . . . , k, and Bu , u = k − i, . . . , 1, in dimension equal to the number of boxes contained in them. This defines an irreducible variety. The strategy is to specialize the flags in order to break such a variety into a union of two varieties that have the same form. The moves on the Mondrian tableaux achieve this purpose. Let M be an admissible Mondrian tableau with an outer box of side length m. If all the D boxes are nested, we slide the smallest A box Ai anti-diagonally up by one unit. Any of the D boxes that touch the lower left hand corner of Ai move one unit up with Ai . The remaining D boxes do not move. If the side length of Ai+1 is not one larger than the side length of Ai or the side length of Bk−i is not 13 m − i (informally, if Ai or Bk−i are not as large as possible given Ai+1 and Bk ), we replace M with the two tableaux described in Possibilities 1 and 2. If the side length of Ai+1 is one larger than the side length of Ai or the side length of Bk−i is m − i, we replace M only with the tableau in Possibility 1. • Possibility 1. We delete Ai and Bk−i and replace them with Di which is the new intersection of Ai and Bk−i . If Di does not intersect or touch Bk−i−1 , we slide all the D boxes anti-diagonally up until Di touches Bk−i−1 . All the remaining boxes stay as in M . • Possibility 2. We shrink the outer box by one so that it passes along the new boundary of Ai and Bk−i and we delete the column and row that lies outside this box. The rest of the boxes stay as in M . Nesting of D boxes Post nesting A6 A5 A4 D3 B6 B2 D4 B1 A4 D3 D2 D1 D2 D1 B1 D3 B1 B2 B1 A5 A4 D3 D3 D2 D1 D3 D2 D1 Figure 4. Admissible Mondrian tableaux and the moves. The two tableaux obtained from M are depicted in Figure 4. Geometrically in the first possibility the k-plane intersects the new intersection Ai ∗ Bk−i . In the second possibility, the k-plane lies in the new span of Ai and Bk−i . Nesting the D boxes. Now suppose that there is an unnested D box Dj . Assume that the smallest square containing Dj−1 and Dj has side length dj . In this case we move Dj−1 anti-diagonally up by one unit. Any D boxes contained in Dj−1 and left justified with it move one unit up with Dj−1 . The remaining boxes stay fixed. If the side length of Dj is less than dj − j + 1 or after the move Dj−1 does not contain Dj , we replace the tableau M with the following two tableaux. If the side length of Dj is dj − j + 1 or after the move Dj−1 contains Dj , we replace M only with the tableau in Possibility 1. • Possibility 1. We delete Dj and Dj−1 . We draw the old span and label it Dj . We also draw the new intersection and label it Dj−1 . If Dj−1 does not meet 14 or touch Ba−i , we slide all the D boxes of index at most j − 1 anti-diagonally up until it does. We keep the remaining boxes as in M . • Possibility 2. We place Dj−1 in its new position and keep all the remaining boxes as in M . It is not hard to check that the results of the moves transform an admissible Mondrian tableau to one or two new admissible Mondrian tableaux. Therefore, we can continue applying the moves to each of the resulting tableaux. After a cycle of moves the number of A and B boxes decrease and the number of nested D boxes increases. Eventually all the boxes will be nested again. The corresponding variety is a Schubert variety. If we apply the moves to each of the Mondrian tableaux that occur until all the boxes are nested, we end up with a collection of tableaux corresponding to Schubert varieties. A dimension calculation shows that applying the degeneration described above to the variety represented by an admissible Mondrian tableau results in the varieties represented by the Mondrian tableaux described in the possibilities. A multiplic­ ity calculation shows that each of the varieties occur with multiplicity one. The following theorem is a consequence of these calculations. Theorem 3.1. The LR coefficient c��,µ of G(k, n) equals the number of times �� results in a game of Mondrian tableaux starting with �� and �µ in an n × n box. We conclude the discussion of the geometric Littlewood - Richardson rules for 2 in G(3, 6) (see the ordinary Grassmannian with an example. We compute �2,1 Figure 5). We start by moving the smallest A box. There are two possibilities. We replace the tableau by the two tableaux where we take the intersection of A 1 and B2 (and slide it up) and keep everything else the same and where we restrict the tableau to the new span of A1 and B2 . We continue resolving the first tableau by moving A2 . Again there are two possibilities. In the second tableau B2 is as large as possible given the outer box, so when we move A1 , there is only one possibility. We then move A2 and now there are two possibilities. We replace the tableau with the tableau where we take the intersection of A2 and B1 and with the tableau where we restrict the tableau to the new span of A2 and B1 . Continuing we conclude that 2 �2,1 = �3,3 + 2�3,2,1 + �2,2,2 . Exercise 3.2. Show that when one takes one of the Schubert cycles to be a special Schubert cycle, one recovers Pieri’s rule. Our proof of the Littlewood - Richardson rule used Pieri’s rule. Carry out the multiplicity calculations explicitly for that case to reprove Pieri’s rule. Exercise 3.3. 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