# An Atlas For Bun r (X)

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1 An Atlas For Bun r (X) As told by Dennis Gaitsgory to Nir Avni October 28, Bun r (X) Is Not Of Finite Type The goal of this lecture is to find a smooth atlas locally of finite type for the stack Bun r (X) of rank-r vector bundles on a smooth projective curve X. Let s see first that it is impossible to find an atlas of finite type: Theorem 1.1. thm:no.global.atlas There is no surjection from a scheme of finite type to Bun r (X). Actually, there are two reasons for this. The first is that the determinant of a vector bundle varies continuously in families, and since there are infinitely many components of Pic(X), it follows that Bun 2 (X) has infinitely many components. On top of that, I ll show now that each connected component is not of finite type. Denote the structure sheaf of X by O. Fix an ample line bundle O(1) on X. For a vector bundle E, we denote E(n) = E O(1) n. If S is a scheme, let X S = X S, and let E S be the pullback of E by the projection X S X. Consider the rank-2 vector bundles O(n) O( n) as points of Bun 2 (X). Theorem 1.1 follows from the following two theorems: Theorem 1.2. thm:connected.sl For every n, there is a connected variety Y (actually, an affine space), a map Y Bun 2 (X), and two points y 0, y 1 Y, such that y 0 is mapped to O O, and y 1 is mapped to O(n) O( n). After constructing an atlas, this means that all the O(n) O( n) s are in the same connected component. Theorem 1.3. thm:infinite.type There is no map Y Bun 2 (X) from a scheme of finite type Y and points y n Y, n = 0, 1, 2,..., such that y n is mapped to O(n) O( n). I am going to use the following theorem of Serre: Theorem 1.4. Let F be a sheaf on a curve X, and let O(1) be an ample line bundle on X. 1. (absolute version) If n is big enough then E(n) is generated by global sections, and H 1 (X, E(n)) = 0. 1

2 2. (relative version) For every scheme S of finite type, if n is big enough, then E S (n) is generated by global sections, and R 1 p E(n) = 0, where p : X S S is the projection. Proof of Theorem 1.2. I want to show first that F = O X O X is an extension of O X ( n) by O X (n). F (n) is globally generated by Serre s theorem. Therefore, there is a never-zero section s Γ(F (n)). This gives a short exact sequence 0 O X F (n) F (n)/o X = L 0. Note that L must be a line bundle. By untwisting, we get 0 O X ( n) F L( n) 0. By looking at the determinant, the last term must be O X (n). Finally, the space of extensions of O X (n) by O X ( n) is the affine space Ext(O X (n), O X ( n)). Proof of Theorem 1.3. A map Y Bun 2 X is a rank 2 vector bundle E on X Y. By Serre, there is n such that E(n) is globally generated, and hence E y is globally generated for all y. But E yn+1 (n) = O( 1) O(2n + 1) is not globally generated. 2 Bun r (X) Is Locally Of Finite Type On the other hand, it turns out that Bun r (X) is an increasing union of Artin stacks of finite type. More precisely, Theorem 2.1. thm:loc.fin.atlas There are open sub-functors U n Bun r (X), varieties Y n of finite type, and smooth surjective maps Y n U n, such that U n are a cover of Bun r (X). As said in the beginning, the degree of the vector bundle is constant in each connected component. Let s fix this degree, and consider only vector bundles of a given degree. For every vector bundle E on X, if n is large enough then E(n) is generated by global sections and H 1 (X, E(n)) = 0. Let U n be the moduli stack of bundles that satisfy these two conditions. More precisely, for every scheme T, let U n (T ) be the full sub-groupoid of Bun r (X) whose objects are vector bundles E on X T that are generated by global sections and such that R 1 p (E(n)) = 0 (p is the projection X T T ). By the theorem of Serre (in the version that applies to families), U n, as n runs over N, cover Bun r (X). Lemma 2.2. U n is an open sub-functor of Bun r (X). Proof. The claim is that, for every scheme S and every vector bundle E on X S, the set of points s in S such that E s (which is the restriction of E to the fiber X {s}) is globally generated and has zero first cohomology, is open. The subset of S for which E s is globally generated is clearly open. The subset of S for which H 1 (E s ) = 0 is the complement of the support of R 1 p E, and hence open. 2

3 Suppose that T is affine and connected, and let E Ob U n (T ). For every t T, the Euler characteristic of the restriction E t (n) of E(n) to X {t} is χ(e t (n)) = dim H 0 (E t (n)) dim H 1 (E t (n)) = dim H 0 (E t (n)). By flatness, χ(e t (n)) is constant. This means that the sheaf p Hom(O T, E(n)) is a vector bundle. To compute its rank, note that, for every t T deg(e t O(n)) = r deg(o(n)) + deg(e t ), and so the dimension of H 0 (E t (n)), which is the rank of p Hom(O T, E(n)), is d = deg(e t (n)) + r(g 1) = r deg(o(n)) + deg(e) + r(g 1). For every T, let Z n (T ) be the groupoid whose objects are pairs (E, φ), where E Ob U n (T ) and φ : O(n) d E is an epimorphism, and the morphisms between (E, φ) and (E, φ ) are maps f : E E such that f φ = φ. Note that Z n is actually an equivalence relation, rather than a groupoid (i.e. the stabilizers of objects are trivial). Lemma 2.3. Z n is representable by an open subscheme of the Quot-scheme. Proof. Let E be a coherent sheaf over X. Recall that the Quot functor sends a scheme S to Quot E/X (S) = {(G, φ) G is a sheaf over X S which is flat over S, and φ : E S G is an epimorphism}. Let F be the following functor: F (S) = {(G, φ) G is a sheaf over X S which is flat over X S, and φ : E S G is an epimorphism}. Z n is an open sub-functor of F, so it is enough to prove that F is representable by an open sub-scheme of Quot. Suppose we have a map S Quot E/X, i.e. a coherent sheaf G over X S and a surjection φ : E S G. We need to show that the set of points s S such that G s is flat over X s is open. Let p : X S S be the projection. Let A X S be the locus where G is non-flat. A is closed, and so S \ p(a), which is the set we are after, is open (note that X, and so p, is proper). If (E, φ) Z ( T ), then φ gives a map between O d T and p (E(n)). Let Y n (T ) Z n (T ) be the collection of such (E, φ) such that this last map is an isomorphism. It is clear that Y n is open in Z n, and hence in the Quot scheme. In particular, it is representable. Finally, we want to show that the natural map Y n U n is a smooth surjection. In fact, Lemma 2.4. The map Y n U n is a GL d -torsor. Proof. Let S U n be an S-valued point. This is a vector bundle E over X S. Lifting this point to Y n is, by definition, choosing an identification between O d S and p (E). This set of identifications is a GL d torsor, as these are vector bundles of degree d. 3

4 3 Level Structure Pick x X. For every n, let (nx) X be the n th infinitesimal neighborhood of x in X, and let Bun (nx) r X be the following functor: Bun (nx) r X(S) = {(E, φ) E is a rank r vector bundle on X S and φ : E (nx)t = O r (nx) T } X Bun (mx) r X is repre- Proposition 3.1. If n > m, then the map Bun (nx) r sentable. Proof. Let S Bun (mx) r X be a map, i.e. a pair consisting of a vector bundle E on X S together with an isomorphism φ : E (mx)s O(mx) r S. Let F be the fiber product. For every scheme T, F(T ) is an equivalence relation, whose objects are triples of 1. A map T S. 2. A vector bundle F on X T and an isomorphism ψ : F (nx)t O r (nx) T. 3. An isomorphism ξ : E T F. such that φ T = ψ (mx)t of ξ (mx)t. Equivalently, the objects are pairs consisting 1. A map T S. 2. An isomorphism ψ : E (nx)t O r (nx) T. such that the restriction of ψ to E (mx)t is equal to φ T. We want to show that this functor is representable. For any k, consider the group scheme GL(E (kx)s ) a group scheme over (kx) S and its restriction of scalars, G k, to S. Denote the kernel of the map G n G m by G n m. The functor F is a torsor over the group scheme G n m, and, therefore, representable. Let p n : Bun (nx) r X Bun r X be the projection. Proposition 3.2. For every n there is N such that p 1 N U n Bun (Nx) r X is a scheme. Proof. Take n = 0 for example (the proof in general is the same). We want to show that, if N is large enough, the functor F (S) = {(E, φ) E U 0 (S), φ : E (Nx)S = O r (Nx) S as O (Nx)S -modules} is representable. Lemma 3.3. If N is large enough, then, for all E U 0, global sections are determined by their restriction to (N x). More precisely, for every S and every E U 0 (S), the restriction map p (E) p (E (Nx)S ) is injective as a bundle map. 4

5 Proof. W.l.o.g we can assume that S is local. If N is large enough, then, by Riemann Roch, dim H 0 (E s ( Nx)) = 0 for every s S, so any global section of E s that vanishes to order N at x must be zero. Suppose that N is as in the lemma. Given (E, φ) F (S), we map it to (E, η), where η is the composition η : p (E) res p (E (Nx)S ) p φ p (O n (Nx) S ) = O Nr S. The range of the map (E, φ) (E, η), which we denote by f, is the following functor: G(S) = {(E, η) E U 0 (S), η : p (E) O Nr S such that OS Nr /η(p (E)) is flat}. The representability of F follows from the following lemmas: Lemma 3.4. G is representable. Proof. G is isomorphic to the product of Y 0 from the last lecture, and the Grassmannian of d-planes in C Nr, twisted by the Isom(O d S, Od S )-torsor Isom(Od S, p (E)). Lemma 3.5. f is a composition of an open embedding and a closed embedding. In particular, it is representable. Proof. It is enough to test the claim on local rings S. In this case, the claim is that, for any vector bundle E U 0 (S) and any inclusion η : p (E) OS Nr, there is at most one isomorphism φ : p (E Nx ) O(Nx) n S inducing η, and the set over which there is such φ is locally closed. The uniqueness is clear, since p (E) generates p (E (Nx)S ) as an O (Nx)S -module. To prove the locally closed condition, note that η is in the image if and only if it factors through the map p (E) p (E Nx ). Since the image of this map generates p (E Nx ) over O Nx, the requirements are 1. η(p (E)) generates O r (Nx) as a module over O Nx. 2. The kernel of η coincide with the kernel of p (E) p (E (Nx). The first condition is open by Nakayama s lemma. clearly closed. The second condition is 5

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