FRAMED HOLOMORPHIC BUNDLES ON RATIONAL SURFACES

FRAMED HOLOMORPHIC BUNDLES ON RATIONAL SURFACES Abstract. We study the moduli space of framed holomorphic bundles of any rank, over the blow-up of CP 2 at q points. For c 2 =, 2 we introduce an open cover of the moduli space and describe its nerve. In the limit when r we use this result to obtain the homotopy type of the moduli spaces. In particular, we compute the cohomology of the moduli spaces.. Introduction Fix a line L CP 2 and let X q denote the blow-up of CP 2 at q points x,..., x q / L. In this paper we will study the moduli space M r k (X q) of equivalence classes of pairs (E, φ), where E is a holomorphic rank r bundle over X q with c = 0 and c 2 = k, holomorphically trivial at L, and φ : E L O r L is a holomorphic trivialization. This moduli space is a special case of the moduli of framed sheaves introduced in [5], []. See also [6]. The interest on these spaces was motivated by the study of moduli spaces of instantons: in [5], [7] it was shown that M r k (X q) is isomorphic as a real analytic space to the moduli space of based charge k SU(r) instantons over a connected sum of q copies of CP 2. Monad descriptions for these spaces were introduced in [], [7], for X 0 = CP 2, [4], [2] for X and [4], [6], for the general case. In this paper we study the moduli spaces using results about bundles on the blow-up of a complex surface (see [9]): analyzing the effect of blowing up on the topology of the moduli space we reduce the study of the moduli spaces over X q to the moduli spaces over X 0, X. Our first result is Theorem.. Let I {,..., q} and write I = #I. I k there is an open set A I M r k (X q) such that () {A I } I =k is an open cover of M r k (X q); (2) A I A J = A I J ; (3) There are homeomorphisms A I M r I (X q ). Then, for each I with For k = this describes M r (X q ) in terms of the moduli spaces over X 0 and X, which are well understood. Hence we can get a cell structure for M r (X q ). For k = 2 we will prove: Theorem.2. There are open sets A i, N ij M r 2(X q ) such that () {A i } i {N ij } i<j is an open cover of M r 2(X q ); (2) There is an open set A M r 2(X q ) such that A i A j = A for i j. (3) For k / {i, j}, N ij A k = N ij A. For different sets {i, j} {k, l}, N ij N kl =. 2000 Mathematics Subject Classification. 4J60 (58B05, 58D27).

2 (4) There are homotopy equivalences A i M r 2(X ) A M r 2(X 0 ) N ij M r (X ) M r (X ) N ij A i M r (X ) M r (X 0 ) N ij A M r (X 0 ) M r (X 0 ) We then apply these results to the study of the rank stable moduli space, which is defined as follows: when r 2 > r, there is a map M r k (X q) M r2 k (X q) induced by taking direct sum with a trivial rank r 2 r bundle. We define the rank stable moduli space as the direct limit M k (X q ) def = lim M r k(x q ). Using the monad descriptions, it was shown in [3], [8] and [2] that () M k (X 0 ) BU(k), M k (X ) BU(k) BU(k) We will prove as a corollary of theorem. that Theorem.3. There is a homotopy equivalence ( q ) M (X q ) BU() BU() Together with the results of [5] and [7], this shows that for a large class of metrics conjecture. in [2] is false. Using theorem.2 we will compute the cohomology of M 2 (X q ): Theorem.4. Let K C Z[x, x 2, x 3, x 4 ] H ( BU() 4 ) be the ideal generated by the product x x 2 and let K A Z[a, k, a 2, k 2 ] H ( BU(2) 2 ) be the ideal generated by k, k 2. Then, as graded modules over Z, we have an isomorphism i= H ( M 2 (X q ) ) Z[a, a 2 ] K q A r q(q ) K C 2 The plan of this paper is as follows: In section 2 we prove theorem. and in section 3 we use it to prove theorem.3. In section 4 we prove theorem.2 and in the next two sections we apply it to prove theorem.4: in section 5 we study the open cover from theorem.2 in the limit when r ; in section 6 we use the spectral sequence associated with this open cover (see [2]) to prove theorem.4. In the appendix we gather some results about the monad constructions of the moduli spaces for q = 0,. This paper is based on results in the author s thesis [20]. 2. An open cover of M r k (X q) Fix a subset I {,..., q} and let π I : X q X I ( I = #I) be the blow up at points x j, j / I. π I induces a map (2) π I : M r k(x I ) M r k(x q ) Let q k. The objective of this section is to prove Theorem 2.. { π I Mr k (X I ) } I =k is an open cover of Mr k (X q). Furthermore π I M r k(x I ) π JM r k(x J ) = π I JM r k(x I J )

BUNDLES ON RATIONAL SURFACES 3 and we have isomorphisms M r k (X I ) π I π I π I Mr k (X I ) From this open cover we can build a spectral sequence converging to H (M r k (X q)). The case k = 2 will be treated in section 7. For the general case see [20], section 4.3. We turn now to the proof of theorem 2.. We begin by proving the last statement: Proposition 2.2. We have isomorphisms M r k (X I ) π I π I π I Mr k (X I ) where π I and π I are inverses of each other. We also have π I M r k(x I ) = {E M r k(x q ) E Li is trivial for i / I} Proof. From theorem 3.2 in [9] it follows that, if a bundle is trivial on the exceptional divisor then it is also trivial on a neighborhood of the exceptional divisor. Hence, a bundle E X on a blow up π : X X is trivial on the exceptional divisor if and only if Ẽ = π π Ẽ. The statement of the proposition follows. Proof of theorem 2.. From proposition 2.2 it follows that π I M r k(x I ) π JM r k(x J ) = π I JM r k(x I J ) To show that M r k (X q) I =k π I Mr k (X k) we only need to show that a bundle E M r k (X q) is trivial in at least q k exceptional lines (q > k). We prove this result by induction in q. Assume E is not trivial in L. Let p : X q X q be the blow up at x and let E = (π E). Then c 2 (E ) < k so we can apply induction. The proof is completed by noting that we cannot have bundles with negative c 2 by Bogomolov inequality for framed bundles (see [5]). Finally we have to show that π I Mr k (X I ) is open. Let H be an ample divisor. Choose N such that H i (E(NH)) = 0 for all E M r k ( X). Then choose M such that π E(NH + ML) is locally free. Consider the function h : M r k ( X) Z defined by h = dim H (E(NH + ML)). Then, from the exact sequence 0 E(NH) E(NH + ML) T 0 (T has support contained in L) we get H 2 (E(NH + ML)) = H 2 (T ) = 0. Now notice that H 0 (E(NH + ML)) H 0 (π E(NH + ML)) and, since by assumption π E(NH + ML) is locally free and π H is ample, for N large enough we get Hence, we get that H i (π E(NH + ML)) = 0 for i > 0 h = χ(π E(NH + ML)) χ(e(nh + ML)) From Riemann-Roch theorem it follows that h = c 2 (E) c 2 (π E ) + f(n, M, c (X)) where f does not depend on E. The result then follows from the upper-semicontinuity of h (see [0], chapter III, theorem 2.8).

4 3. The charge one moduli space The objective of this section is to prove theorem.3: Theorem 3.. There is a homotopy equivalence ( q ) M (X q ) BU() BU() From theorem 2. we know that q M r (X q ) = πl M r (X ), πi M r (X ) πj M r (X ) = π Mr (X 0 ) (i j) l= i= We begin by studying the maps π Mr (X 0 ) π i Mr (X ). Lemma 3.2. Let ι : CP r CP r CP r be the inclusion into the first factor: ι ([u]) = ([u], ), where denotes the base point. Then there are homotopy equivalences h 0 : CP M (X 0 ) and h : CP CP M (X ) such that the following diagram π M (X 0 ) π M (X 0 ) h 0 CP is homotopy commutative. π i π π i M (X ) M (X ) h ι CP CP Proof. We will use the monad description of M r (X ), M r (X 0 ) (see appendix A). We define the following maps: p 0 : M r (X 0 ) CP r p : M r (X ) CP r CP r : CP r CP r CP r p 0 : [a, a 2, b, c] [b] ( [ ]) c t p : [a, a 2, d, b, c] [b], c 2 : [u] ([u], [u]) f : CP CP CP CP f : (x, y) (x, xy ) where to define f we observe that CP = BU() is homotopic to the free abelian group on U(). Now observe that the diagram π M (X 0 ) π π i M (X 0 ) π π i M (X ) M (X ) p 0 CP ι p CP CP f CP CP is homotopy commutative and the maps p 0, p, f, π, π i are homotopy equivalences. The statement of the lemma then follows by writing h 0 = p 0 and h = p f, where p 0, p, f are the homotopy inverses. We are ready to prove theorem 3..

BUNDLES ON RATIONAL SURFACES 5 Proof of theorem 3.. Let C be the cone on q points v,..., v q. Let ( q BU() BU() {v i }) ( BU() C ) M = i= ( [u], v i ) ( ι ([u]), v i ) We first show that M is homotopically equivalent to M (X q ). Denote the points in C by [t, v i ] C = [0, ] i {v i} (0, v i ) (0, v j ) Define a map ( q ) ζ : BU() BU() {v i } ( BU() C ) M (X q ) i= as follows: denote the points in C by [t, v i ] C = [0,] i {vi} (0,v i) (0,v j). Then define ζ by BU() BU() {v i } ([u ], [u 2 ], v i ) π i h ([u ], [u 2 ]) BU() C ([u], [t, v i ]) π h 0([u]) for t < 3 BU() C ([u], [t, v i ]) πi h ι ([u]) for t > 2 3 For 3 t 2 3 use the homotopy between π h 0 and p i h ι from lemma 3.2. ζ descends to the quotient to give a map ζ : M M (X q ). We want to apply Whitehead theorem to show ζ is a homotopy equivalence. The van Kampen theorem implies both M and M (X q ) are simply connected hence we only have to show ζ is an isomorphism in homology groups. We prove it by induction in q =,..., q. We apply the five lemma to the Meyer-Vietoris long exact sequence corresponding to open neighborhoods of the sets π q +M (X ), π (,...,q ) M (X q ) M (X q ) BU() BU() {v i }, BU() BU() {v l } M using the fact that the restrictions q l= ζ : BU() BU() {v i } π i M (X ) ζ : BU() C π M (X 0 ) are homotopy equivalences. It follows that ζ induces isomorphisms in all homology groups. To conclude the proof we only have to show that M is homotopically equivalent to q ( q ) BU() BU() {v l } l= BU() BU() = (x,, v i ) (x,, v j ) i= where BU() is the base point. Define an open cover of BU() ( i BU()) by U i = BU() BU() {v i }. Then the claim is a special case of proposition 4. in [2].

6 4. An open cover of M r 2(X 2 ) The objective of this section is to prove theorem.2. We begin by studying the case q = 2. We will adopt, in this section and the next, the following notation: Denote the blow up points by x L, x R X 0. Let π : X 2 X 0 be the blow up map at x L, x R. By abuse of notation we will denote by π L the maps X 2 X and X X 0 corresponding to the blow up at x L and in the same way π R will denote the blow up at x R. We have the diagram π L X 2 X R πr π L X L X 0 π R of blow up maps where X L XR X. Denote by L L and L R the exceptional divisors above x L and x R respectively. Again, by abuse of notation we identify L L X 2 with L L X L and the same for L R. Write x L = [x L, x 2L, ], x R = [x R, x 2R, ], x L, x R X 0 = CP 2. Since x L x R we may assume without loss of generality that x L x R. Let z i = x ir x il. z, z 2 determine a point ([z, z 2, ], [z, z 2 ]) X \ L = CP 2 \ L CP 2 CP. We are ready to state the first theorem of this section: Theorem 4.. Let A L = π RM r 2(X L ) = {E M r 2(X 2 ) : E LR A R = π LM r 2(X R ) = {E M r 2(X 2 ) : E LL is trivial} is trivial} and let C = M r 2(X 2 ) \ (A L A R ). Let N L M r 2(X L ) be the set of non-degenerate configurations m = (a, a 2, d, b, c) such that the eigenvalues of da (equal to the eigenvalues of a d) are in a δ neighborhood of 0, z. In a similar way define N R M r 2(X R ). Let N 2 = π R N L π L N R C. Then {A L, A R, N 2 } is an open cover of M r 2(X 2 ). There are homotopy equivalences () A L A R M r 2(X ) (2) C M r (X ) M r (X ) (3) A L A R M r 2(X 0 ) (4) A L N 2 N L A R N 2 N R M r (X ) M r (X 0 ) (5) A L A R N 2 M r (X 0 ) M r (X 0 ) (6) N 2 C From this open cover we get, in a standard way (see [2]), a spectral sequence: Corollary 4.2. There is a spectral sequence converging to the cohomology of M r 2(X 2 ) with E term E 0,n = H n (A L ) H n (A R ) H n (N 2 ) E,n = H n (A L A R ) H n (A L N 2 ) H n (A R N 2 ) E 2,n = H n (A L A R N 2 ) In the next section we will study the d differential of this spectral sequence. We turn now to the proof of theorem 4.. We will delay the proof that N 2 is open and begin by proving the homotopy equivalences (), (2) and (3):

BUNDLES ON RATIONAL SURFACES 7 Proposition 4.3. A L, A R are open sets, C = {[E, φ] M r 2(X 2 ) : c 2 ( (π i E) ) =, i = L, R} and the following maps are isomorphisms (where πi def (E) = (π i E) ): π R : M r 2(X L ) A L M r 2(X 2 ) π L : M r 2(X R ) A R M r 2(X 2 ) π R π L : C S 0 M r (X L ) S 0 M r (X R ) π : M r 2(X 0 ) A L A R M r 2(X 2 ) where S 0 M r (X ) M r (X ) is the subspace of bundles E verifying (π E) = O r X 0. Proof. The isomorphisms for A L, A R, A L A R follows from theorem 2.. That theorem also implies A L, A R are open. It remains to look at the map πr πl : C S 0M r (X L ) S 0 M r (X R ). The continuity of this map was proved in proposition 3. in [9]. We will construct an inverse for πr π L. Let (E L, φ L ) S 0 M r (X L ), (E R, φ R ) S 0 M r (X R ). Hartogs theorem implies there are unique extensions of φ L, φ R to maps φ L : E L X0\{x L } O r X 0\{x L }, φ R : E R X0\{x R } O r X 0\{x R } These maps induce an isomorphism E L ER over X 0 \ {x L, x R } which we use to glue E L, E R and obtain a bundle E X 2. The continuity of this map was proved in proposition 3.3 in [9]. This concludes the proof. Before we continue we need a lemma. Let M r 2 (X ) be the Donaldson-Uhlenbeck completion of the moduli space M r 2(X ) (see the appendix). A blow-up π : X 2 X induces a map π : M r 2(X 2 ) M r 2 (X ) given by E ((π E), l((π E) /π E)) Lemma 4.4. Let E m M r 2 (X ) and let m = (a, a 2, d, b, c) be the configuration associated to E m. The following are equivalent: () E m is in the image of π R : C M r 2 (X ); (2) cdb = 0 and the eigenvalues of da i (equal to the ones of a i d) are 0 and z i ; (3) After a change of basis we can write a = [ ] a 0 0 z, a 2 = with c b = 0. [ a 2 b c z b c z z 2 ], d = [ ] 0 0 0 Proof. We will show that 2, 2 3 and 3. [ ] b, b =, c = [ c c ] 2: Suppose E m = π R Ẽ for some Ẽ C. Then, by proposition 4.3, Em S 0 M r (X L ) and E m is not locally free at the blow up point x R. So, from proposition A.7, Em corresponds to a configuration of the form [a, a 2, 0, b, c ]. Since m is degenerate, by proposition A.5 after a change of basis it can be written in one of two forms, corresponding to the two types of special pairs. If m is b-special then [ ] [ ] [ ] a a i = i d b 0 a, d = i 0 d, b =, c = [ c 0 c ] b

8 in which case the configuration is equivalent to the completely reducible configuration (see proposition A.5) (a, a 2, d, b, c ) (a, a 2, d, 0, c ) (3) (4) corresponding to an ideal bundle with singularity at (a d, a 2d ) and charge one bundle given by (a, a 2, d, b, c ). So we should have d = 0 and a i d = z i. Hence the eigenvalues of da i are 0, z i and cdb = 0. A similar argument applies if m is c-special. 2 3: Now assume the configuration (a, a 2, d, b, c) satisfies 2. Fix a basis of eigenvectors v 0, v V of a d and w 0, w W of da with v 0, w 0 corresponding to the eigenvalue 0. Normalize v, w so that dv = w. Then [ ] [ ] a a = 0 a b c 2 0 a a, a 2 = d a b c d a a a 2 [ ] [ ] d 0 b d =, b = 0 b, c = [ c c ] From cdb = 0 we get (b c )(b c ) = 0. If b c = 0 then a 2 is lower triangular. If b c = 0 then a 2 is upper triangular. In both cases the diagonal entries of a 2 d are its eigenvalues. Hence, the condition about the eigenvalues of a d and a 2 d yields the equations a d = a 2d = 0, a = z and a 2 = z 2. Since a (W ) + a 2 (W ) + b(c r ) = V we must have d = 0. 3 : Let m = [a, a 2, d, b, c] be a configuration satisfying 3. c b = 0 implies either c = 0 or b = 0. It follows that the pair (Span{(0, )}, Span{(0, )}) is a special pair hence the configuration is degenerate. Now, from proposition A.5 it follows that m is equivalent to the completely reducible configuration m m = (a, a 2, 0, b, c ) (z, z 2, 0, 0) Notice that (a, a 2, 0, b, c ) S 0 M r (X L ). Then, from proposition 4.3, there is m C such that π R m = m. Then, from the characterization of points in the completion it follows that π R m = m. Now we turn to the proof of homotopy equivalences (4) and (5) in theorem 4.. Definition 4.. Let N z = { (a z, a 2z, b z, c z ) M r (X 0 ) a z z < δ } N = { (a, a 2, d, b, c ) M r (X ) d a < δ } Let N 0 M r 2(X 0 ) be the subset of points (a, a 2, b, c) with the eigenvalues of a lying in δ neighborhoods of x L and x R. Then define the map 0 : N xl N xr N 0 by [a L, a 2L, b L, c L ] 0 [a R, a 2R, b R, c R ] = [a, a 2, b, c] with a = [ ] al 0 0 a R, a 2 = [ a 2L b R c L a L a R b L c R a R a L a 2R ], b = [ bl ], c = [ ] c b L c R R

BUNDLES ON RATIONAL SURFACES 9 and L : N N z N L by [a, a 2, d, b, c ] L [a, a 2, b, c ] = [a, a 2, d, b, c] with [ ] [ ] a a = 0 a b c 2 0 a a, a 2 = d a (5) b c d a a a 2 [ ] [ ] d 0 b d =, b = 0 b, c = [ c c ] Proposition 4.5. () The maps 0, L are homeomorphisms; (2) The inclusions N z M r (X 0 ), N M r (X ) are homotopy equivalences; (3) π R N L π L N R = π N 0. Proof. Statement (2) is clear from the definition. To prove statement (3) we observe that π RN L π LN R = π RN L π Mr 2(X 0 ) N L π LM r 2(X 0 ) The result now follows easily from proposition A.8. We turn to the proof of statement (). It is an easy consequence of proposition A.5 that 0 and L preserve the nondegeneracy of the configurations so the maps are well defined. Now we look at L. For δ small enough the eigenvalues of a d are distinct. Hence we can choose, up to the action of (C ) 4, eigenvector basis {v 0, v } V of a d and {w 0, w } W of da, where v 0, w 0 correspond to the eigenvalues near 0. Normalize v, w so that dv = w. Then the action of (C ) 4 is reduced to an action of (C ) 3. We can thus write (see also equation (3)) [ ] [ ] a 0 a b c 2 (6) a = 0 a, a 2 = [ ] d 0 d =, b = 0 b c d a a [ b b ] a d a a 2, c = [ c c ] The group (C ) 3 acts transitively on equivalence classes of such configurations written in the above canonical form. This shows the existence of an inverse, hence L is a homeomorphism. The proof for 0 is similar. We will need the following identity: Proposition 4.6. Let τ : M r k (X 0) M r k (X 0) be defined by τ(a, a 2, b, c) = (a x L, a 2 x 2L, b, c) Let m, m 2 M r (X 0 ). Then π L (m 0 m 2 ) = π L m L τ(m 2 ). Proof. It follows easily from proposition A.8. The maps 0, L extend to the closure N, N z of N, N z. The following proposition is a direct consequence of proposition A.5: Proposition 4.7. Let m L = [a L, a 2L, b L, c L ] N xl, m R = [a R, a 2R, b R, c R ] N xr. Then the following are equivalent: () m L 0 m R is degenerate; (2) Either m L or m R is degenerate. (3) At least one of the 4 vectors b L, b R, c L, c R is zero.

0 Let m = [a, a 2, d, b, c ] N, m = [a, a 2, b, c ] N z. The following are equivalent: () m L m is degenerate; (2) Either m or m is degenerate; (3) One of the 4 vectors b, b, c, c is zero. We are ready to prove Proposition 4.8. N 2 is an open neighborhood of C. Proof. From lemma 4.4 it follows immediately that π R C N L. Suppose there is a sequence y n M r 2(X L ) such that y n y π R C. Write y n = [a n, a 2n, d n, b n, c n ]. Then, by property 2 in lemma 4.4 the eigenvalues of d n a in converge to 0, z i. Hence, for n large enough y n N L. Hence N L π R C is an open neighborhood of π R C. Suppose there is a sequence x n x C such that x n / N 2. Hence x n / C so, by passing to a subsequence we may assume without loss of generality that x n πr Mr 2(X L ). Let y n = π R x n M r 2(X L ) and write y n = [a n, a 2n, d n, b n, c n ]. Then y n y = π R x by continuity of π R, and y n / N L. But by property 2 in lemma 4.4 the eigenvalues of d n a in converge to 0, z i which implies, for n large enough, that y n N L. Finaly we prove the homotopy equivalence (6): Proposition 4.9. The inclusion C N 2 is a strong deformation retract. Proof. We will construct a homotopy H 2 : N 2 [0, ] N 2 betwen the identity and a retraction N 2 C. Let H x,x 2 : Nz [0, ] N z be defined by H x,x 2 (a, a 2, b, c, t) = ( t 2 a + ( t 2 )x, t 2 a 2 + ( t 2 )x 2, tb, tc ) and let H : N [0, ] N be defined by H (a, a 2, d, b, c, t) = (a, a 2, t 2 d, b, c ) Then we defined H L : NL [0, ] N L by H L (m L m, t) def = H (m, t) L H z,z 2 (m, t) We define H 2 as the unique solution of the system of equations (7) π R H 2 (x, t) = H L (π R x, t) π L H 2 (x, t) = H R (π L x, t) We have to show existence and uniqueness of solution. Then we will show that H 2 defines a homotopy between the identity on N 2 and a retraction N 2 C. We define the auxiliary map H 0 : N0 [0, ] N 0 by H 0 (m L 0 m R, t) def = H xl,x 2L (m L, t) 0 H xr,x 2R (m R, t) To prove existence and uniqueness of solution of the system (7) we consider two cases: () Assume that either t = 0 or x C. Then we claim that H L (π R x, t) π R C, H R (π L x, t) π L C. If t = 0 this follows directly from lemma 4.4. If x C then, from lemma 4.4 we can write π R x = x L x = (a, a 2, 0, b, c ) L (a, a 2, b, c )

BUNDLES ON RATIONAL SURFACES with c b = 0. It then follows from the definition of H L that H L (π R x, t) = π R x for all t. In the same way we see that H R (π L x, t) = π L x. This proves the claim. Then, existence and uniqueness follows from proposition 4.3. (2) Assume t and x / C. Then we may assume π R x N L. Then, since H L (π R x, t) N L, we get from (7) π R H 2 (x, t) = H L (π R x, t) H 2 (x, t) = π RH L (π R x, t) This proves uniqueness. To prove existence we need to show that π L H 2 (x, t) = π L π RH L (π R x, t) = H R (π L x, t) It is enough to show this for the case where x = πl π Ry for some y N 0 since the set of points of this form is dense and H L, H R, π L, πr, π R are continuous. It is an easy computation to show that H L (πl y, t) = πl H 0(y, t), H R (πr y, t) = π R H 0(y, t). It follows that π L π RH L (π R x, t) = π L π Rπ LH 0 (y, t) = π RH 0 (y, t) = H R (π L x, t) Now we need to show that H 2 is the desired homotopy. Direct inspection shows H 2 (x, ) = x. We saw in () above that, for x C, H 2 (x, t) = x and H 2 (x, 0) C. The continuity of H 2 follows from the continuity of π L, π R, H L, H R : let (x n, t n ) (x, t) N 2. Then, after passing to a subsequence, we get H 2 (x nk, t nk ) ( x, t ) N 2. Then equations 7 imply that ( x, t ) N 2 and unicity of solution implies ( x, t ) = H 2 (x, t). Applying this reasoning to every sublimit of H 2 (x n, t n ) we conclude that H 2 (x n, t n ) H 2 (x, t). Hence H 2 is continuous. We now prove the general case (theorem.2). What remains to be proven is: Theorem 4.0. Let A i, A be as in theorem 2. and let N ij = π ij N 2. Then () {A i } {N ij } is an open cover of M r 2(X q ); (2) There are homotopy equivalences N ij A i M r (X ) M r (X 0 ) N ij A M r (X 0 ) M r (X 0 ) (3) For k / {i, j}, N ij A k = N ij A ; (4) For different sets {i, j} {k, l}, N ij N kl =. Proof. Statement () follows from theorems 2. and 4.. Statement (2) follows from proposition 2.2. To prove statement (3) observe that N ij A k = N ij A ij A k. So we turn to statement (4). First we look at N ij N jk. Let x,..., x q CP 2 be the blow-up points and write x i = [x i, x 2i, ]. We may assume without loss of generality that the balls B δ (x i ) are disjoint. Now observe that N ij N jk = N ij A ij A jk N jk = N ij A j N jk Let z i = x i x j and let z k = x k x j. Then, theorem 4. states that N ij A j is the set of non-degenerate configurations (a, a 2, d, b, c) such that the eigenvalues of da are in a δ neighborhood of 0 and z i. Hence N ij A j and N jk A j are disjoint. This shows N ij N jk =. The case N ij N kl for {i, j} {k, l} = is treated similarly.

2 5. The differential d The objective of this section is to obtain, in the limit when r, the homotopy type of the inclusion maps (see theorem 4.) A L A 0 N L A R N 0 N R N 2 where A 0 = A L A R = π Mr 2(X 0 ). Since these spaces are classifying spaces it is enough to study the pullback under these maps of the tautological bundles. This will allow us to compute the d differential in the spectral sequence introduced in corolary 4.2, from where we will compute the homology of M 2 (X 2 ). Definition 5.. Let { (u, v) Hom(C F k, C r ) Hom(C k, C r ) u, v are injective } 0 (k, r) = (u, v) (u(ḡ t ), vg), g Gl(k, C) { (u, v) Hom(C F k, C r ) Hom(C k, C r ) u, v are injective } (k, r) = (u, v) (u(ḡu) t, vg v ), g u, g v Gl(k, C) and define F 0 (k, r) F 0 (k, r) and F (k, r) F (k, r) by { [ bt F 0 (k, r) =, c ] F } 0 (k, r), bc = 0 { [ bt F (k, r) =, c ] F } (k, r), bc = 0 Lemma 5.. Let j 0 : F 0 (k, r) M r k (X 0), j : F (k, r) M r k (X ) be the inclusion maps given by [ bt, c ] [b, c]. Then we have the homotopy commutative diagram (8) M r k (X 0) π j 0 j p 0 F 0 (k, r) ı F0 0 (k, r) M r k (X ı ) F (k, r) F (k, r) pr pr p 0 Gr(k, C r ) where p 0 is the projection [b, c] [c]. Moreover, in the rank stable limit, the maps ı 0, ı, p 0, p 0, j 0, j are homotopy equivalences. Proof. We divide the proof into three steps: () p 0, p 0 are fibrations with fibers M(k, r k) and M(k, r) respectively where M(k, r) = U(r) U(k) is the space of injective maps from Ck to C r which is contractible in the stable range. That proves p 0, p 0 are homotopy equivalences. It imediatelly follows that ı 0 is a homotopy equivalence. (2) Now we look at ı. Consider the projection p : F (k, r) Gr(k, C r ) given by [b, c] [c]. When r, the spectral sequence associated with the fibration p : F (k, r) Gr(k, C r ) (whose fiber is Gr(k, C r k )) collapses since all homology is in even dimensions. It easilly follows that ı is an isomorphism in all homology groups, hence an homotopy equivalence.

BUNDLES ON RATIONAL SURFACES 3 (3) Finally we need to prove the statements about j 0, j. First we look at j 0. Let R be the space of configurations (a, a 2, b, c) and let R F R be the subspace configurations of the form (0, 0, b, c). Then we have the fibration map F 0 (k, r) R F j 0 R M r k (X 0) In the rank stable limit the spaces R F and R are contractible (see [8], [2]) so, by the five lemma j 0 is an isomorphism in homotopy groups hence an homotopy equivalence. A similar proof works for j. Now we turn to the main theorem of this section. Definition 5.2. Let E, L be the tautological bundles over Gr(2, ) and Gr(, ) respectivelly. Consider the compositions (9) A 0 π M 2 (X 0 ) j 0 F 0 (2, ) p 0 Gr(2, C ) (0) () N 0 N 0 0 N xl N xr p L N xl M (X 0 ) p0j 0 Gr(, C ) 0 N xl N xr p R N xr M (X 0 ) p0j 0 Gr(, C ) (2) N L L N N z p N z M (X 0 ) p0j 0 Gr(, C ) Then we define the following bundles: E 0 A 0 is the pullback of E under the composition 9. L 0L,0 N 0 is the pullback of L under 0 L 0R,0 N 0 is the pullback of L under L 0R,L N L is the pullback of L under 2 Now let Ẽu, Ẽv F (2, r) be the tautological bundles corresponding to u, v and let L u, L v be the tautological line bundles over F (, ). Consider the compositions (3) A L π R M 2 (X L ) j F (2, ) ı F (2, ) (4) N L L N N z p N M (X ) ıj F (, ) (5) N 2 C π L S 0 M (X L ) M (X L ) ıj F (, ) (6) N 2 C π R S 0 M (X R ) M (X R ) ıj F (, ) We define the bundles E bl, E cl A L are the pullback of Ẽu, Ẽv under 3. L bl,l, L cl,l N L are the pullback of L u, L v under 4 L bl,2, L cl,2 N 2 are the pullback of L u, L v under 5 L br,2, L cr,2 N 2 are the pullback of L u, L v under 6 Theorem 5.2. We have the following bundle isomorphisms:

4 () E bl A0 = E 0, E cl A0 = E 0 (2) L bl,l N0 L0L,0, L cl,l N0 L0L,0, L 0R,L N0 L0R,0 (3) E bl NL LbL L 0R, E cl NL LcL L 0R (4) E 0 N0 L0L,0 L 0R,0. (5) L br,2 NL LcR,2 NL L0R,L (6) L bl,2 NL LbL,L, L cl,2 NL LcL,L Similar statements hold for the spaces A R, N R and the maps N R A R, N R N 2 and N 0 N R. Proof. () First we show that E bl A0 EcL A0 E0. Consider diagram (8). We will start by defining a homotopy inverse q : Gr(k, C r ) F 0 (k, r) to the map p 0 : F 0 Gr(k, C ) as follows: choose a map c : C k C r representing an element [c] Gr(k, C r ). Choose h Gl(k, C) such that ch is orthogonal. Then define q([c]) = [ch, ch]. This map is well defined and independent of the choice of h. Also p 0 q = hence p 0 = q. Now observe that the composition pr q : Gr(k, C r ) F (k, r) = Gr(k, C r ) Gr(k, C r ) is the diagonal map. It follows that, if E is the tautological bundle over Gr(k, C ), then q pr Ẽ u q pr Ẽ v E To show that E bl A0 EcL A0 E0 it suffices to show that pr ı Ẽu pr ı Ẽv p 0E. We have pr ı Ẽu = ı 0 pr Ẽ u p 0 q pr Ẽ u = p 0E and a similar statement is true for Ẽv. This concludes the proof. (2) We want to show that L bl,l N0 L0L,0, L cl,l N0 L0L,0 and L 0R,L N0 L 0R,0. We have the commutative diagram (see proposition 4.6) N 0 0 N xl N xr p R N xr M (X 0 ) j 0 F 0 p 0 Gr N L π L π τ N N z τ p N z M j 0 (X 0 ) F 0 τ p 0 Gr from which it follows that L 0R,L N0 L0R,0. We also have the commutative diagram N 0 0 N xl N xr p L N xl M (X 0 ) j 0 F 0 ı 0 p 0 F 0 p 0 Gr N L π L π τ N N z τ τ p N M j ı (X ) F F pr pr from which it follows, as in step (), that L bl,l N0 LcL,L N0 L0L,0.

BUNDLES ON RATIONAL SURFACES 5 (3) We want to show that E bl NL LbL,L L 0R,L and E cl NL LcL,L L 0R,L. Consider the following diagram: (7) j A L F N L L N N z M (X ) M (X 0 ) F F 0 ı j j 0 ı ı 0 F w F F 0 Since pr Lu pr Lv p 0 L, the proof will be complete if we show there is a map w : F (, ) F 0 (, C ) F (2, ) making the diagram homotopy commutative, such that (8) w Ẽ u = L u pr Lu, w Ẽ v = L v pr Lv We begin by building w. Define maps s L, s R : Gr(, C ) Gr(, C ) as follows: let v : C C and write v = (v, v 2,...). Then s L ([v]) def = [(v, 0, v 2, 0,...)], s R ([v]) def = [(0, v, 0, v 2,...)] We observe that s L, s R are homotopic to the identity. It follows that, if we define then w : ([b L, c L ], [b R, c R ]) [s L (b L ) s R (b R ), s L (c L ) s R (c R )] ( w) Ẽ u = L u pr Lu, ( w) Ẽ v = L v pr Lv It remain to show diagram 7 is commutative. Let j z : F 0 (, ) N z be defined by j z : [b, c] [z, 0, b, c]. Then the diagram j j z N N z F F 0 M (X ) M (X 0 ) is homotopy commutative. We are left with the diagram j A L F ı F w N L L j j z ı ı 0 N N F z F 0 F F 0 Now define the map w : F (, ) F 0 (, ) F (2, ) by w : ([b L, c L ], [b R, c R ]) [s L (b L ) s R (b R ), s L (c L ) s R (c R )] Clearly we have the commutative diagram F (2, ) w ı F (2, ) w F (, ) F 0 (, ) ı ı 0 F (, ) F 0 (, )

6 We are thus left with the diagram j A L F N L L Next we introduce maps j j z N N F z F 0 S L ([b, c]) = [s L (b), s L (c)], S R ([b, c]) = [s R (b), s R (c)] These maps are homotopic to the identity hence we only have to show the diagram j A L F N L L j j z N N F z F 0 F F 0 w S L S R is homotopy commutative. This is an easy direct verification. (4) We want to show that E 0 N0 L0L L 0R. Consider the diagram w i N 0 A 0 i 2 i3 i 4 N L A L Then E 0 = i 3E bl so E 0 N0 = i E 0 = i i 3E bl = i 2i 4E bl = L 0L L 0R (5) We want to show that L br NL LcR NL L0R. The result will follow if we show that the following diagram is homotopy commutative: L (9) N L N N N M z z (X 0 ) F 0 F0 N 2 C π L S 0 M (X ) π M (X ) F F pr Let S L N 2 = {(E, φ) N 2 c 2 ((π L E) ) = } S N L = {(E, φ) N L c 2 ((π L E) ) = } S 0 N = {(E, φ) N c 2 ((π L E) ) = 0}

BUNDLES ON RATIONAL SURFACES 7 Then the commutativity of (9) follows from the commutativity of (20) N 2 N L π R C S L N 2 S N L π π L R π π L L M π R (X R ) M (X 0 ) L L N N z S 0 N N z We need to check the image of L : S 0 N N z N L is contained in S N L. Then, analyzing the commutativity of diagram (20) boils down do analyzing the diagram N z (2) S N L L S 0 N N z π L M (X 0 ) Let m S 0 N S 0 M (X ), m = [a, a 2, 0, b, c ]. Let m N z. Then a direct computation shows that (π L (m L m )) = m. This shows that the image of S 0 N N z under L is contained in S N L and that diagram (2) is commutative. (6) We want to show that L bl,2 NL LbL,L, L cl,2 NL LcL,L. This will follow from the commutativity of the diagram L N z (22) N L N N N M z (X ) N 2 π R C π R S 0 M (X ) M (X ) We showed in proposition 4.9 that the map H 2 (, 0) is the homotopy inverse of the inclusion C N 2. Let (m, m ) N N z. Then, by definition of H 2, π R H 2 (π R(m L m ), 0) = H L (m L m, 0) = H (m, 0) H z (m, 0) Hence the diagram N L π R L N N N z N 2 H 2(,0) C π H (,0) R S 0 M (X ) is commutative. From here it follows easily that diagram (22) is commutative. 6. The cohomology of M 2 (X q ) The objective of this section is to prove theorem.4. We begin by proving it for the special case q = 2:

8 Theorem 6.. There is an exact sequence 0 H (M 2 (X 2 )) H (A L ) H (A R ) H (N 2 ) which splits to give an isomorphism H (A 0 ) H (N L ) H (N R ) H (N 0 ) 0 H (M 2 (X 2 )) Ker ( H (A L ) H (A R ) H (A 0 ) ) Ker ( H (N 2 ) H (N L ) H (N R ) ) Proof. Recall corolary 4.2. We will use this spectral sequence to compute H (M 2 (X 2 )). Clearly the map d : E,n E 2,n is surjective hence E 2,n 2 = 0. Also we notice that E p,2n+ = 0 for any p. It follows that the spectral sequence collapses at the term E 2. We get then ( ) (23) H 2n ( M 2 (X 2 ) ) = E 0,2n = Ker d : E 0,2n E,2n ( ) Ker d : E,2n H 2n+ ( M 2 (X 2 ) ) = E,2n E 2,2n (24) = ( ) Im d : E 0,2n E,2n When performing calculations we will use the following sign conventions: + (25) A L A 0 N L + + N 0 + N 2 + A R + N R We begin by defining the following generators of the cohomology of E 0,2n : a i L = c i (E cl ) c i (E bl ) a i R = c i (E cr ) c i (E br ) a i bl = c i (E bl ) a i br = c i (E br ) c L = c (L cl ) c (L bl ) c R = c (L cr ) c (L br ) c bl = c (L bl ) c br = c (L br ) We do the same for E,2n : n L = c (L cl ) c (L bl ) n R = c (L cr ) c (L br ) n bl = c (L bl ) n br = c (L br ) n 0R = c (L 0R ) n 0L = c (L 0L ) a i = c i (E 0 ) and for E 2,n : n 0R = c (L 0R ) n 0L = c (L 0L ) Then, from theorem 5.2 it follows that the map d : E 0,2n E,2n may be represented by the following diagram, whose entries correspond to those in diagram

BUNDLES ON RATIONAL SURFACES 9 (25): ( a L, a bl, a2 L, ) a2 bl ( n L, n bl n 0R, n L n 0R, n bl n 0R ) (n L, n bl, 0, n 0R ) ( 0, a, 0, a 2) ( a R, a br, a2 R, a2 br) (c L, c bl, c R, c br ) Also the map d : E,2n E 2,2n is given by Now let Then (a, a 2 ) (n 0L + n 0R, n 0L n 0R ) (n L, n bl, n 0R ) (0, n 0L, n 0R ) (n 0L, n R, n br ) (n 0L, 0, n 0R ) ( n R, n br n 0L, n R n 0L, n br n 0L ) (0, n 0L, n R, n br ) K AL = Ker(H (A L ) H (A 0 )) K AR = Ker(H (A R ) H (A 0 )) K NL = Ker(H (N L ) H (N 0 )) K NR = Ker(H (N R ) H (N 0 )) H (A L ) Z[a, a 2 ] K AL, H (A R ) Z[a, a 2 ] K AR H (C) Z[n L, n R ] K NL K NR K C H (N L ) Z[n L, n R ] K NL, H (N R ) Z[n L, n R ] K NR Notice that K C H (C) is the ideal generated by c L c R. The restriction of the map H (A L ) H (N L ) to K AL induces a map s L : K AL K NL. Similarly we have a map s R : K AR K NR. Let also s : Z[a, a 2 ] Z[n L, n R ] be the map induced by the direct sum map BU() BU() BU(2). Then the map d : E 0,2n E,2n is given by d (a L + k AL, a R + k AR, x + k NL + k NR + k C ) = = ( s(a L ) s L (k AL ) + x + k NL, s(a R ) s R (k AR ) x k NR, a L + a R ) and the map d : E,2n E 2,2n is given by Now we can finish the proof: d (x L + k NL, x R + k NR, a) = (x L + x R + s(a)) () We prove first that H 2n+ (M 2 (X 2 )) = 0. We need to show Ker (d : E,2n E 2,2n ) Im (d : E 0,2n E,2n ). Let (x L + k NL, x R + k NR, a) Ker d. Then x L + x R + s(a) = 0. It follows that d (a, 0, x R + k L k R ) = (x L + k L, x R + k R, a) (2) Now we will show that H 2n Z[a, a 2 ] K AL K AR K C which conpletes the proof. We first define a map Z[a, a 2 ] K AL K AR K C E 0,2n by (a, k AL, k AR, k C ) (a + k AL, a + k AR, s(a) + s L (k AL ) s R (k AR ) + k C )

20 We want to show this map is injective onto the kernel of d. Injectivity is clear and a direct verification shows the image is contained in the kernel of d. To show surjectivity let (a L + k AL, a R + k AR, x + k NL + k NR + k C ) Ker d. Then a L = a R, k NL = s L (k AL ), k NR = s R (k AR ) and x = s(a L ) = s(a R ). The result follows. We are ready to prove the general case: Theorem 6.2. With notations as in theorem 2. let K i = Ker ( H (π i M 2 (X )) H (π M 2 (X 0 )) ) K ij = Ker ( H (π ijm 2 (X 2 )) H (π i M 2 (X )) H (π j M 2 (X )) ) Then, as modules over Z, we have an isomorphism (26) H ( M 2 (X q ) ) H ( M 2 (X 0 ) ) i K i i<j K ij Proof. We divide the proof into two steps: () We will use theorem 2. to build a spectral sequence converging to the cohomology of H (M 2 (X q ). Let be the q simplex. Label its vertices by v i, i =,..., q, and the e ij be the middle point of the edge joining v i and v j. We define a filtration 0 of where 0 = i<j e ij and is the -skeleton of. Write = i i where i is the closure of the connected component of \ 0 containing v i. Then we define M = i,j ( eij π ij M 2 (X 2 ) ) i ( i π i M 2 (X )) ( π M 2 (X 0 ) ) where is induced by the inclusions e ij i and π M 2 (X 0 ) πi M 2 (X ) πij M 2 (X 2 ). Then, the arguments in [2] can be applied to show that M is homotopically equivalent to M 2 (X q ). The filtration of by 0, induces a filtration F 0 F F 2 = M of M which leads to a spectral sequence with E 0,n = H n (F 0 ) i<j ( H 0 (e ij ) H n (π ijm 2 (X 2 )) ) ( H ( i, i ) H n (π i M 2 (X )) ) E,n = H n (F, F 0 ) i E 2,n = H n (F 2, F ) H (F ) H n (M 2 (X 0 )) (2) The d differential is induced by the inclusions π M 2 (X 0 ) π i M 2 (X ) π ij M 2 (X 2 ). We will use the sign conventions (i < j): (27) H ( πij M 2 (X 2 ) ) + H ( πi M 2 (X ) ) H ( π j M 2 (X ) ) + H ( π M 2 (X 0 ) ) +

BUNDLES ON RATIONAL SURFACES 2 Let K i = Ker ( H (π i M 2 (X )) H (π M 2 (X 0 )) ) K ij = Ker ( H (π ijm 2 (X 2 )) H (π i M 2 (X )) H (π j M 2 (X )) ) Then, from theorem 6. we have H (π i M 2 (X )) H (π M 2 (X 0 )) K i H (π ijm 2 (X 2 )) H (π M 2 (X 0 )) K i K j K ij Then the sequence of maps E 0,n d E,n d E 2,n splits into three sequences (28) i<j H0 (e ij) K ij 0 0 i<j H0 (e ij ) (K i K j ) l H ( l, l ) K l 0 H 0 ( 0 ) K n H (, 0 ) K n H ( ) K n (29) where K n stands for H n (M 2 (X 0 )). The bottom maps are easily analyzed using the exact sequence 0 H 0 ( ) H 0 ( 0 ) H (, 0 ) H ( ) 0 It follows that the map d : E,n E 2,n is surjective. Since E r,n = 0 for r > 2 and n even, this implies the spectral sequence collapses and H n (M 2 (X q )) = Ker(d : E,n E 2,n ) Im(d : E 0,n E,n ) H n (M 2 (X q )) = Ker(d : E 0,n E,n ) Lets look more closely at the map H ( i, i ) K i Observe that i<j H 0 (e ij )(K i K j ) i i<j H 0 (e ij ) ( K i K j ) = i H 0 ( i ) K i It follows that the map (29) can be easily analysed using the exact sequence 0 H 0 ( i ) H 0 ( i ) H ( i, i ) 0 We gather together our conclusions: (a) The top sequence in (28) contributes a term H 0 (e ij ) K ij i<j to H 2n (M 2 (X q )). (b) The bottom sequence in (28) does not contribute to H 2n+ (M 2 (X q )) since it is exact in the middle. (c) The bottom sequence in (28) contributes a term to H 2n (M 2 (X q )). H 0 ( ) H (M 2 (X 0 ))

22 (d) The map (29) is surjective hence it does not contribute to H 2n+ (M 2 (X q )). (e) The map (29) contributes a term H 0 ( i ) K i i to H 2n (M 2 (X q )). From (b) and (d) it follows that H 2n+ (M 2 (X q )) = 0 and from (a), (c) and (e) equation (26) follows. Appendix A. Monads In this appendix we will sketch the monad description of the spaces M r k (CP2 ) and M r k ( CP 2 ). We follow [2]. See also [3]. Let L CP 2 be a rational curve and let L CP 2 be the exceptional divisor. Choose sections x, x 2, x 3 spanning H 0 (O(L )) and y, y 2 spanning H 0 (O(L L)) so that x 3 vanishes on L and x y + x 2 y 2 spans the kernel of H 0 (O(L )) H 0 (O(L L)) H 0 (O(2L L)) A.. The moduli space over CP 2. Let W be a k-dimensional vector space. Let R be the space of 4-tuples m = (a, a 2, b, c) with a i End(W ), b Hom(C r, W ), c Hom(W, C r ), obeying the integrability condition [a, a 2 ] + bc = 0. For each m = (a, a 2, b, c) R we define maps A m, B m by A m = W ( L ) x a x 3 x 2 a 2 x 3 cx 3 A m W 2 C n Bm W (L ), B m = [ x 2 + a 2 x 3 x a x 3 bx 3 ] Then B m A m = 0. The assignement m E m = Ker B m /Im A m induces a map f : R M r k (CP2 ). m is called non degenerate if A m, B m have maximal rank at every point in CP 2. Theorem A.. f induces an isomorphism betwen the quotient of the space of non degenerate points in R by the action of Gl(W ): and the moduli space M r k (CP2 ). For a proof see [7], proposition. g (a, a 2, b, c) = (g a g, g a 2 g, g b, cg) Theorem A.2. The algebraic quotient R/Gl(W ) is isomorphic to the Donaldson- Uhlenbeck completion of the moduli space of instantons over S 4. For a proof see [8], sections 3.3, 3.4, 3.4.4. We sketch here how the map from R/Gl(W ) to the Donaldson-Uhlenbeck completion of the moduli space of instantons is constructed (see [2] for details): Let m = (a, a 2, b, c) R. A subspace W W is called b-special with respect to m if (30) a i (W ) W (i =, 2) and Im b W. A subspace W W is called c-special with respect to m if (3) a i (W ) W (i =, 2) and W Ker c.

BUNDLES ON RATIONAL SURFACES 23 m is called completely reducible if for every W W which is b-special or c-special, there is a complement W W which is c-special or b-special respectively. Proposition A.3. Let m = (a, a 2, b, c) R. () m is non degenerate if and only if the only b-special subspace is W and the only c-special subspace is 0; (2) For every m, the orbit of m under Gl(W ) contains in its closure a canonical completely reducible orbit and completely reducible orbits have disjoint closures; (3) If m is completely reducible then, after acting with some g Gl(W ) we can write [ ] [ ] a red a i = i 0 b red 0 a, b =, c = [ c i 0 red 0 ] where (a red, a red 2, b red, c red ) is non-degenerate and the matrices a, a 2 can be simultaneously diagonalized. Such a configuration is equivalent to the following data: An irreducible integrable configuration (a red, a red 2, b red, c red ) corresponding to a bundle with c 2 = l k; k l points in C 2 = CP 2 \ L given by the eigenvalue pairs of a, a 2 This is precisely the Donaldson-Uhlenbeck completion. A.2. The moduli space over CP 2. Let R be the space of 5-tuples m = (a, a 2, d, b, c) where a i Hom(W, V ), d Hom(V, W ), b Hom(C r, V ), c Hom(W, C r ), such that a (W ) + a 2 (W ) + b(c r ) = V, obeying the integrability condition a da 2 a 2 da + bc = 0. For each m = (a, a 2, d, b, c) R we define maps A m, B m by W ( L ) V (L L ) A m (V W ) 2 C n B m V (L ) W (L L) A m = a x 3 y 2 x da x 3 0 a 2 x 3 y x 2 da 2 x 3 0 cx 3 0, B m = [ x2 a 2 x 3 x a x 3 bx 3 dy y dy 2 y 2 0 Then B m A m = 0. The assignement m E m = Ker B m /Im A m induces a map f : R M r k ( CP 2 ). A point m R is called non-degenerate if A m and B m have maximal rank at every point in CP 2. Theorem A.4. The map f induces an isomorphism betwen the quotient of the space of non degenerate points in R by the action of Gl(V ) Gl(W ): (g 0, g ) (a, a 2, b, c, d) = (g 0 a g, g 0 a 2g, g 0 b, cg, g dg 0) and the moduli space M r k ( CP 2 ). See [2] for a proof. Consider the algebraic quotient R/Gl(V ) Gl(W ). This space is a completion of the moduli space M r k ( CP 2 ). We proceed to give an interpretation of the points in this completion in terms of the Donaldson-Uhlenbeck completion. See [2] for ]

24 details. Let m = (a, a 2, d, b, c). Let V V and W W and assume dim V = dim W. The pair (V, W ) is called b-special with respect to m if (32) a i (W ) V (i =, 2), d(v ) W and Im b V The pair (V, W ) is called c-special with respect to m if (33) a i (W ) V (i =, 2), d(v ) W and W Ker c m is called completely reducible if for every pair (V, W ) which is either b-special or c-special, there are complements V, W to V and W such that the pair (V, W ) is c-special or b-special respectively. Proposition A.5. Let m = (a, a 2, d, b, c) R. () m is non-degenerate if and only if the only b-special pair is (V, W ) and the only c-special pair is (0, 0); (2) For every m, the orbit of m under Gl(V ) Gl(W ) contains in its closure a canonical completely reducible orbit and completely reducible orbits have disjoint closures; (3) If m is completely reducible then, after acting with some (g 0, g ) Gl(V ) Gl(W ), we can write [ ] [ ] [ ] a red a i = i 0 d red 0 b red 0 a, d = i 0 d b =, c = [ c 0 red 0 ] where (a red, a red 2, d red, b red, c red ) is non-degenerate effective and integrable and the matrices a, a 2, d can be simultaneously diagonalized. Such a configuration is equivalent to the following data: An irreducible configuration (a red, a red 2, d, b red, c red ) associated to a bundle with c 2 = l k; k l points in the blow up C 2 of C 2 at the origin. This points are determined as follows: a, a 2, d determine k l unique points (λ r, λ r 2), [µ r, µ r 2] C 2 corresponding to vectors v,..., v k l such that da i v r = λ r i vr (λ, λ 2 are the eigenvalue pairs of da, da 2 ) and (µ r a + µ r 2a 2 )v r = 0. A.3. Direct image. In this section we gather some results concerning the direct 2 image map π induced by the blowup map π : CP CP 2. Proposition A.6. Let π # : R R be given by π # (a, a 2, d, b, c) = (da, da 2, db, c). Let m R, m = π # m. Then E m CP 2 \L is isomorphic to E m CP 2 \[0,0,]. For the proof see [9], proposition 5.6. Proposition A.7. Let S 0 M r ( CP 2 ) = {(E, φ) M r (CP 2 ) : (π E) = O r CP 2 }. Then () m S 0 M r (CP 2 ) if and only if m is of the form (a, a 2, 0, b, c). (2) The inclusion S 0 M r M r is a homotopy equivalence. Proof. First we observe that M r ( CP 2 ) = S 0 M r ( CP 2 ) π Mr (CP 2 ). Now m π Mr (CP 2 ) if and only if d is an isomorphism (see [2]). The first statement follows. The second statement follows easily from the first: just consider the homotopy (a, a 2, d, b, c) (a, a 2, td, b, c).

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26 20., Topology of moduli spaces of rank stable instantons and holomorphic bundles, Ph.D. thesis, Stanford University, 2002. 2. Graeme Segal, Classifying spaces and spectral sequences, Inst. Hautes Études Sci. Publ. Math. (968), no. 34, 05 2. MR 38 #78 Instituto Superior Técnico E-mail address: jsantos@math.ist.utl.pt