Topics in Geometry: Mirror Symmetry

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1 MIT OpenCourseWare Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit:

2 MIRROR SYMMETRY: LECTURE 8 DENIS AUROUX Last time: Linear Algebra. Today: Multivariable Calculus. / 18.0 Complex Variables Thursday: Differential Equations 1. Mirror Symmetry Conjecture Last time, we said that if we have a large complex structure limit (LCSL) degeneration, then we have a special basis (α 0,..., α S, β 0,..., β S ) of H 3 (X, Z) s.t. β 0 is invariant under monodromy and β 1,..., β s are mapped by monodromy φ j by β i βi m ji β 0 for m ji Z. We decided that we would normalize so that φ j β 0 Ω = 1, and let w i = β i Ω (w i w i m ji ) and q i = exp(2πiw i ) (which we called canonical coordinates). ( ) Example. Given a family of tori T with monodromy, Ω = 1, 0 1 Ω = a b τ (precisely what you get identifying the elliptic curve with R 2 /Z τz), q = exp(2πiτ). If e i is a basis of H 2 ( X, ˇ Z), e i in the K ahler cone, we obtain coordinates on the complexified Kähler moduli space: if [B + iω] = ť i e i, let ˇq i = exp(2πiť i ), ť i = B + iω. e i Example. Returning to our example, ˇq = exp(2πi T 2 B + iω). Conjecture 1 (Mirror Symmetry). Let f : X (D ) S be a family of Calabi- Yau 3-folds with LCSL at 0. Then a Calabi-Yau 3-fold Xˇ and choices of bases α 0,..., α S, β 0,..., β S of H 3 (X, Z), e 1,..., e S of H 2 (X, Z) s.t. under the map m : (D ) S M Kah (Xˇ), (q 1,..., q S ) (ˇq i,..., qˇs ), qˇi = q i, we have a coincidence of Yukawa couplings (1),, X,, Xˇ qi q j q p = m(p) k qˇi qˇj qˇk where the LHS corresponds to Ω ( Ω) and the RHS to a (1, 1) X qi qj qk Xˇ coupling, i.e. the Gromov-Witten invariants e i, e j, e k 0,β (since 2πiqˇi qˇi = ťi = e i H 1,1 etc.). 1

3 2 DENIS AUROUX Remark. A more grown-up version of mirror symmetry would give you an equivalence between H (X, T X) with its usual product structure and H ( ˇX, C) with the quantum twisted product structure as Frobenius algebras (making this concrete would require more work) Application to the Quintic (See Gross-Huybrechts-Joyce, after Candelas-de la Ossa-Greene-Parkes). Last time, we defined (2) X ψ = {(x 0 : : x ) P f ψ = with (3) G = {(a = (Z/Z) 3 0,..., a ) (Z/Z) a i = 0}/{(a, a, a, a, a)} so it is LCSL. We want to compute the periods of the holomorphic volume form on Xˇψ. There is a volume form Ωˇ ψ on Xˇψ induced by the G-invariant volume form Ω ψ on X ψ by pullback via π : Xˇψ X ψ /G. We want to find a 3-cycle β 0 H 3 (Xˇψ) that survives the degeneration. For z = 0, { x i = 0} contains tori in component P 3 s, e.g. 0 x ψx 0 x 1 x 2 x 3 x = 0} i acting by diagonal multiplication x i x i ξ a i, ξ = e 2πi/. We obtained a crepant resolution Xˇψ of X ψ /G (its singularities are C ij = {x i = x j = 0}/G), which has h 1,1 = 101, h 2,1 = 1, and h 3 =. The map (x 0 :... : x ) (ξ a x 0 : x 1 :... : x ) gives X ψ = X ξφ, so let z = (ξ). Then z 0, i.e. ψ, gives a toric degeneration of X ψ to {x 0 x 1 x 2 x 3 x = 0}. This is maximally unipotent, as the monodromy on H 3 is given by () () T 0 = {(x 0 : : x ) x = 1, x 0 = x 1 = x 2 = δ, x 3 = 0} We want to extend it to z 0. Take x = 1, x 0 = x 1 = x 2 = δ: then x 3 should be given by the root of f ψ which tends to 0 as ψ. We need to show that there is only one such value (giving us a simple degeneration rather than a branched covering). Explicitly, set x 3 = (ψx 0 x 1 x 2 ) 1/ y: (6) f ψ = 0 x 0 + x 1 + x 2 + (ψx 0 x 1 x 2 ) / y + 1 (ψx 0 x 1 x 2 ) / y i.e. y x 0 + x 1 + x (7) y = + (ψx 0 x 1 x 2 ) /

4 MIRROR SYMMETRY: LECTURE 8 3 One root is y ψ / 0, with the other four roots converging to. So for x 3, we have one root ψ 1, and roots ψ 1/. Now, G acts freely on T 0 X ψ, and T 0 /G is contained in the smooth part of X ψ /G and gives a torus Ť 0 Xˇψ, β 0 = [Ť 0 ]. Because T 0, Ť 0 still make sense at z = 0, their class is preserved by the monodromy. Next, we want to get the required holomorphic volume form. In the affine subset x = 1, let Ω ψ be the 3-form on X ψ characterized uniquely by (8) Ω ψ df ψ = ψdx 0 dx 1 dx 2 dx 3 f ψ at each point of X ψ. At a point where x3 = 0, ( x 0, x 1, x 2 ) are local coordinates, and ψdx 0 dx 1 dx 2 ψdx 0 dx 1 dx 2 (9) Ω ψ = f ψ = x3 ψx 0 x 1 x 2 x 3 Defining it in terms of other coordinates, we get the same formula on restrictions. We need to extend this to where x = 0. We could rewrite this using homogeneous coordinates, but note that the corresponding divisor is just the canonical divisor: since X ψ is Calabi-Yau, this divisor has no zeroes or poles at x = 0. Since Ω ψ is G-invariant, it induces a 3-form on (X ψ /G) nonsing and lifts and extends to Ωˇ ψ on Xˇψ with 1 (10) Ωˇ ψ = Ω ψ Tˇ 3 0 T 0 Tool: we have the residue formula 1 (11) f(z)dz = res f (z i ) 2πi S 1 z i poles of f D 2 So let T = { x 0 = x 1 = x 2 = x 3 = δ, x = 1}. Then ( ) 1 ψdx 0 dx 1 dx 2 dx 3 1 ψdx 3 (12) = dx 0 dx 1 dx 2 2πi T f ψ T 2πi 3 S f x 1 ψ 0 x 1 x 2 where f ψ has a unique pole at x 3. The residue is precisely ψ (13) = dx 0 dx 1 dx 2 = Ω ψ T 0 ( f/ x 3 ) T 0 ψ ( f/ x 3, giving us )

5 DENIS AUROUX So 1 dx 0 dx 1 dx 2 dx 3 Ω ψ = 2πi (ψ) 1 (x 0 + x 1 + x 2 + x 3 + 1) x 0 x 1 x 2 x 3 T0 T (1) = 1 ( dx 0 dx 1 dx 2 dx 3 1 (ψ) 1 x 0 + x 1 + x 2 + x ) 1 2πi T x 0x 1 x 2 x 3 x 0 x 1 x 2 x 3 1 dx 0 dx 1 dx 2 dx 3 (x 0 + x 1 + x 2 + x 3 + 1) m = 2πi T x 0x 1 x 2 x 3 (ψ) m (x 0 x 1 x 2 x 3 ) m We want to find the coefficient of 1 in the latter term. We obviously need m = n (the numerator only has powers which are a multiple of ), and want the coefficient of x n x n x n x n in (x 3 + 1) n, which is (n)! x 1 + x 2 + x. We (n!) finally obtain (1) Ω ψ = (2πi) 3 (n)! T (n!) 0 (ψ) n In terms of z = (ψ), the period is proportional to (n)! (16) φ n 0 (z) = z (n!) Set a n = (n)! (n!). Then (17) (n + 1) (n + )! a n+1 = = (n + )(n + 3)(n + 2)(n + 1)a n (n!) (n + 1) d Setting Θ = z dz : Θ( c n z n ) = nc n z n, giving us the Picard-Fuchs equation (18) Θ φ 0 = z(θ + 1)(Θ + 2)(Θ + 3)(Θ + )φ 0 Next time, we will show that there is a unique regular solution, and a unique solution with logarithmic poles to our original problem.