[ Pobierz całość w formacie PDF ]
.4.QEDBy de nition, the tangent space of M at the point A0; 0 is the quotientKer T =Im G : in fact we consider the linear approximation to the SeibergWitten equations modulo those directions that are spanned by the action ofthe gauge group.Thus we need to compute H1 C to get the virtual tangentspace.The index theorem provides a way to compute the Euler characteristicof C in terms of some characteristic classes.Theorem 4.4 The Euler characteristic of C is, C = Ind DA + d+ + d ;where d is the adjoint of the exterior derivative.Proof: Up to zero order operators G can be deformed to the exterior derivatived; and T can be deformed to the pair of operatorsDA : , X; W+ ! , X; W,andd+ : 1 ! 2+:Hence the assembled complex becomesDA +d+ +d0 ! 1 , X; W+ ! 0 2+ , X; W, ! 0:The Euler characteristic of the original complex is not a ected by this change,hence:, C = Ind DA + d+ + d :QEDpNote that, since L is not really a line bundle, its rst Chern class is de nedpc1 L1to be c1 L = , and it makes sense in the coe cient ring Z.Z2 242 Corollary 4.5 The index of the complex C is equal top2 +32c1 L , ;4p2where is the Euler characteristic of X, is the signature of X, and c1 L ispthe cup product with itself of the rst Chern class of L integrated over X withp p2 2a standard abuse of notation we write c1 L instead of c1 L ; X.Proof: Use the additivity of the index.By the index theorem for the twistedDirac operator it is known thatZp^Ind DA = , ch L A X :Xp p1 2^The Chern character is ch L = 2 1 + c1 L + c1 L + the rank of L21^ ^over the reals is two.The A class is A X = 1 , p1 X + , where p1 X24is the rst Pontrjagin class of the tangent bundle.Thus the top degree termp1 2^of ch L A X will be p1 X + c1 L.On the other hand, by the index12 R1theorem for the signature operator it is known that p1 X = , hence we3 Xgetp2Ind DA = c1 L , :4The index of d+ + d can be read o from the chain complexd d+0 ! 0 ! 1 ! 2+ ! 0:The Euler characteristic of this complex turns out to be1,Ind d + d+ = + ;2by another application of the index theorem.Summing up together it follows thatp2 +32C = c1 L , :4QEDNow, in order to obtain from this index computation the dimension of themoduli space, we need the following lemmata:Lemma 4.6 The tangent space TA; M at a regular point =0 can be iden-6ti ed with H1 C.43 Proof: H1 C describes exactly those directions that are spanned in nitesimallyat the point A; by the solutions of the Seiberg Witten equations, modulothose directions that are spanned by the action of the gauge group.QEDNotice that in the following, unless otherwise speci ed, we use the notationM to denote the moduli space corresponding to a xed choice of the Spincstructure in S, and for a generic choice of the metric and of the perturbation.Sometimes the dependence on the Spinc structure is stressed by adding a sub-script Ms, with s 2 S.Lemma 4.7 In the above complex H0 C =0 and, under a suitable perturba-tion of the Seiberg Witten equations, also H2 C = 0.Proof: H0 C = 0 since the map G, which describes the in nitesimal actionof the gauge group, as in 17 , is injective.In order to show that H2 C = 0,we use the following strategy.We allow the perturbation to vary in 2+, andconsider at a given point A; ; the operator1~TA; ; ; ; = DA + i 0; d+ + , Im eiej 0; ei ^ ej ;2that maps the L2 -tangent space to the L2 one.Here A; is a solution ofk k,1the perturbed equations as in de nition 3.6, with perturbation.We prove that~the operator T + G , with~T+G0 ! 1 2+ , X; W+ ! 2+ , X; W+ 0 ! 0;is surjective.This implies, by the in nite dimensional Sard theorem, that, for ageneric choice of the perturbation , the original operator T + G is surjective,hence H2 C = 0.~The operator T has closed range.Suppose there is a section in 2+ , S+~L that is orthogonal to the image of T in the space of L2 sections.We wantk, 1to show that such an element must be identically zero.By an argument of~21 , this is enough to prove that T is surjective on smooth sections as well.Soassume that ; is orthogonal to any section of the form1DA + i 0; d+ + , Im eiej 0; ei ^ ej :2This means thatDA0 + i 0; =0;with the inner product of sections of W, , and1d+ + , Im eiej 0; ei ^ ej; =0244 ~as 2 forms.Using the fact that ; is in the kernel of T , which is an ellipticoperator with L2 coe cients, we can assume that ; is in fact in L2 [ Pobierz całość w formacie PDF ]

  • zanotowane.pl
  • doc.pisz.pl
  • pdf.pisz.pl
  • gieldaklubu.keep.pl
  •