MIT OpenCourseWare http://ocw.mit.edu 8.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
MIRROR SYMMETRY: LECTURE 3 DENIS AUROU Last time, we say that a deformation of (, J) is given by () {s Ω 0, (, T ) s + 2 [s, s] = 0}/Diff() To first order, these are determined by Def (, J) = H (, T ), but extending these to higher order is obstructed by elements of H 2 (, T ). In the Calabi-Yau case, recall that: Theorem (Bogomolov-Tian-Todorov). For a compact Calabi-Yau (Ω n,0 = O ) with H 0 (, T ) = 0 (automorphisms are discrete), deformations of are unobstructed. Note that, if is a Calabi-Yau manifold, we have a natural isomorphism T = Ω n, v i v Ω, so (2) H 0 (, T ) = H n,0 () = H 0, and similarly (3) H (, T ) = H n,, H 2 (, T ) = H n,2. Hodge theory Given a Kähler metric, we have a Hodge operator and L 2 -adjoints (4) d = d, = and Laplacians (5) Δ = dd + d d, = + Every (d/)-cohomology class contains a unique harmonic form, and one can show that = Δ. We obtain 2 (6) H k (, C) dr = Ker (Δ : Ω k (, C) ) = Ker ( : Ω k ) = Ker ( : Ω p,q ) = H p,q () p+q=k p+q=k
2 DENIS AUROU The Hodge operator gives an isomorphism H p,q = H n p,n q. Complex conjugation gives H p,q = H q,p, giving us a Hodge diamond h n,n h n,n h 0,n h n,n h n,n.... (7)............. h, h 0,.... For a Calabi-Yau, we have h n,0 h,0 h 0,0 (8) H p,0 = H n,n p = H n p (, Ω n ) = H n p (, O = H n p,0 ) = H 0,n p Specifically, for a Calabi-Yau 3-fold with h,0 = 0, we have a reduced Hodge diamond 0 0 (9) 0 h, h 2, 0 0 h 2, h, 0 0 0 Mirror symmetry says that there is another Calabi-Yau manifold whose Hodge diamond is the mirror image (or 90 degree rotation) of this one. There is another interpretation of the Kodaira-Spencer map H (, T ) = H n,. For = (, J t ) t S a family of complex deformations of (, J), c (K ) = c (T ) = 0 implies that Ω n (,J = under the assumption H O t) () = 0, so we don t have to worry about deforming outside the Calabi-Yau case. Then [Ω t ] H n,0 J () H n (, C). How does this depend on t? Given T 0 S, t t t Ω t Ω n,0 Ω n, by Griffiths transversality: (0) α t Ω p,q = α t Ω p,q + Ω p,q+ + Ω p+,q J t t
MIRROR SYMMETRY: LECTURE 3 3 Since Ωt t t=0 is d-closed (dω t = 0), ( Ωt t=0 ) (n,) is -closed, while t () ( Ω t t=0 ) (n,) + ( Ω t t=0 ) (n,) = 0 t t Thus, [( Ωt t=0 ) (n,) ] H n, (). t For fixed Ω 0, this is independent of the choice of Ω t. If we rescale f(t)ω t, f Ω (2) t (f(t)ω t ) = Ω t + f(t) t t t Taking t 0, the former term is (n, 0), while for the latter, f(0) scales linearly with Ω 0. (3) H n, () = H (, Ω n ) = H (, T ) and the two maps T 0 S H n, (), H (, T ) agree. Hence, for θ H (, T ) a first-order deformation of complex structure, θ Ω H (, Ω n T ) = H n, () and (the Gauss-Manin connection) [ θ Ω] (n,) H n, () are the same. We can iterate this to the third-order derivative: on a Calabi-Yau threefold, we have (4) θ, θ 2, θ 3 = Ω (θ θ 2 θ 3 Ω) = Ω ( θ θ2 θ3 Ω) where the latter wedge is of a (3, 0) and a (0, 3) form. 2. Pseudoholomorphic curves (reference: McDuff-Salamon) Let ( 2n, ω) be a symplectic manifold, J a compatible almost-complex structure, ω(, J ) the associated Riemannian metric. Furthermore, let (Σ, j) be a Riemann surface of genus g, z,..., z k Σ market points. There is a well-defined moduli space M g,k = {(Σ, j, z,..., z k )} modulo biholomorphisms of complex dimension 3g 3 + k (note that M 0,3 = {pt}). Definition. u : Σ is a J-holomorphic map if J du = du J, i.e. J u = (du + Jduj) = 0. For β H 2 (, Z), we obtain an associated moduli 2 space (5) M g,k (, J, β) = {(Σ, j, z,..., z k ), u : Σ u [Σ] = β, J u = 0}/ where is the equivalence given by φ below. u Σ, z,..., z k (6) φ = u Σ, z,..., z k This space is the zero set of the section J of E Map(Σ, ) β M g,k, where E is the (Banach) bundle defined by E u = W r,p (Σ, Ω 0, u T ). Σ
4 DENIS AUROU We can define a linearized operator D : W r+,p (Σ, u T ) T M g,k W r,p (Σ, Ω 0, U T ) Σ (7) D (v, j ) = 2 ( v + J vj + ( v J) du j + J du j ) = v + 2 ( v J)du j + 2 J du j This operator is Fredholm, with real index (8) index R D := 2d = 2 c (T ), β + n(2 2g) + (6g 6 + 2k) One can ask about transversality, i.e. whether we can ensure that D is onto at every solution. We say that u is regular if this is true at u: if so, M g,k (, J, β) is smooth of dimension 2d. Definition 2. We say that a map Σ is simple (or somewhere injective ) if z Σ s.t. du(z) = 0 and u (u(z)) = {z}. Note that otherwise u will factor through a covering Σ Σ. We set M g,k (, J, β) to be the moduli space of such simple curves. Theorem 2. Let J (, ω) be the set of compatible almost-complex structures on : then (9) J reg (, β) = {J J (, ω) every simple J-holomorphic curve in class β is regular} is a Baire subset in J (, ω), and for J J reg (, β), M (, J, β) is smooth g,k (as an orbifold, if M g,k is an orbifold) of real dimension 2d and carries a natural orientation. The main idea here is to view J u = 0 as an equation on Map(Σ, ) M g,k J (, ω) (u, j, J). Then D is easily seen to be surjective for simple maps. We have a universal moduli space MM π J J (, ω) given by a Fredholm map, and by Sard-Smale, a generic J is a regular value of π J. This universal moduli space is M = J J (,ω) M g,k(, J, β). For such J, M g,k(, J, β) is smooth of dimension 2d, and the tangent space is Ker (D ). For the orientability, we need an orientation on Ker (D ). If J is integrable, the D is C-linear (D = ), so Ker is a C-vector space. Moreover, J 0, J J reg (, β), a (dense set of choices of) path {J t } t [0,] s.t. t [0,] M g,k(, J t, β) is a smooth oriented cobordism. We still need compactness.