1. What is this course about?2. Preliminaries about the Grassmannian3. A Littlewood-Richardson ruleReferencesGRASSMANNIANS: 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. 1The 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. 2Additional 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 Pn−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 = Cn , 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