Symplectic Topology. Ivan Smith

Transcription

1 Symplectic Topology Ivan Smith Lent 2006

2 L1: Introduction Riemannian geometry is the geometry of a nondegenerate symmetric bilinear form: we equip each tangent space T x M of a manifold M with such a pairing. Symplectic geometry is the geometry of a non-degenerate skew-symmetric bilinear form, which moreover we insist is locally constant as we vary from tangent space to tangent space. Formally, fix the skew form (( ) ( )) Ω = diag on R 2n, i.e. this is the 2n 2n matrix which in 2 2 blocks just has these equal diagonal entries. The group Sp 2n (R ) GL 2n (R ) is the subgroup of matrices which preserve this form: A t ΩA = Ω. A symplectic manifold is a 2nmanifold with an atlas of charts for which the transition maps are diffeomorphisms of R 2n whose derivatives belong to Sp 2n (R ).

3 Examples: Euclidean space R 2n is trivially symplectic, and is the local model for all symplectic manifolds: there are no local invariants, in contrast to curvature in Riemannian geometry. Open subsets and products of symplectic manifolds inherit symplectic structures. Motivation: If a particle moves in R n with force Φ = U/ q, for a potential U, Newton s law is: q = U/ q. Letting H = U + q 2 /2 be total energy and p = q, Hamilton s equations: H/ p = q; H/ q = ṗ for (q, p) R 2n the position and momentum. We are used to conservation of energy: d dt H(p, q) = H q q + H p ṗ = 0 But the flow also preserves the symplectic structure of R 2n ; time evolution in classical mechanics is a map R Symp(R 2n ).

4 A symplectic matrix has determinant +1, so symplectic manifolds have canonically defined volume forms: measure the volume of a subset by covering it in patches and using the measure on R 2n. The choice of charts doesn t matter since changing charts U α φ αβ U β multiplies the local integrals by det(dφ αβ ) = 1. Theorem (Moser): (i) two such volume forms on a closed manifold are equivalent if and only if they have the same total volume; (ii) if U, V are open subsets of R k then there is a volumepreserving embedding U V if and only if vol(u) vol(v ). Theorem (Gromov): There is no symplectic embedding B 2n (R) B 2 (r) R 2n 2 if R > r. This is called the non-squeezing theorem. This amounts to saying symplectic geometry can t be reduced to volume-preserving geometry, and will come at the end of the course.

5 Motivations from symmetry: suppose we try and classify groups acting locally on R k for which (i) the group acts locally transitively (or we could just reduce dimension to an orbit) (ii) the group has no invariant foliation : it s not of the form (x, y) φ(x, y) = (f(x), g(x, y)) for R k = R l R k l (or simplify by φ f). Theorem (Lie): If such a symmetry group is finite-dimensional and compact it s SO(n), or SU(n) or SP (n) or one of a finite list of exceptions. [Drop compactness: get SO(p, q), SO(n, C ) etc, but still finitely many families.] Theorem (Cartan): If the symmetry group is infinite-dimensional it s Diff(R k ) or Vol(R k ), or Symp(R 2k ) or Cont(R 2k+1 ), or a conformal analogue, with no exceptional cases. Symplectic geometry, and it s odd-dimensional cousin contact geometry, are hence very natural, but we won t come back to this.

6 Motivations from topology: four-dimensional geometry is very special. For instance, R 2n has a unique smooth structure if n 2, but R 4 has uncountably many. In the world of closed simply-connected manifolds, in a given homeomorphism type in dimension k 4 there are at most finitely many diffeomorphism types and (complicated) homotopy invariants distinguish these but in dimension 4 there are often infinitely many diffeomorphism types, and no known homotopy interpretation of how to separate them. Conjecture (Donaldson): If X, Y are homeomorphic symplectic four-manifolds, then X is diffeomorphic to Y if and only if X S 2 and Y S 2 are deformation equivalent as symplectic manifolds. i.e. symplectic geometry in dimension 6 should at least, often does capture exotic smooth geometry in dimension 4.

7 Quantum cohomology: a symplectic manifold has a natural deformation of the ring structure in H (X, R ). Recall the intersection pairing in cohomology is non-degenerate so a cohomology class A is completely determined by giving A, c Z for every c. If now A is itself a cup-product A = a b then we see that the three-point function (a, b, c) a b, c Z completely determines the cup-product. Think of this via intersections of cycles: We re counting intersection points, i.e. maps P 1 X which are constant and have image inside C a C b C c. Generalise: count nonconstant holomorphic maps P 1 X which take 0 C a, 1 C b and C c. We think of this as a deformation: count maps with area N as terms of order ε N in an expansion...

8 Plan of the course: Part 1: Background differential forms on real and complex manifolds, almost complex structures, the first Chern class. Part 2: Moser methods basic symplectic geometry. What do symplectic manifolds look like locally? Which manifolds are symplectic What are special symplectic submanifolds? Part 3: Constructions how to build symplectic manifolds by surgery, and take quotients by group actions, and so forth. We ll state without proof a topological characterisation of symplectic manifolds. Part 4: Holomorphic curves invariants of symplectic manifolds, by counting certain minimal surfaces inside them. We ll prove the non-squeezing theorem, deduce a topological characterisation of symplectic diffeomorphisms, and define quantum cohomology.

9 L2: Differential forms We start with linear algebra. For a vector space E, the tensor algebra T (E) is i 0 E i (where E 0 = F is the underlying field). The tensor product is defined by saying there is a bilinear map E E E E, and the RHS is universal for this: all bilinear maps from E E factor through it. If v i, v i are bases for E and E then v i v j is a basis for E E. Quotient by the ideal v v to skew-symmetrise and obtain the exterior algebra Λ (E); Λ k E is universal for alternating multilinear maps E k U; equivalently Λ k E = E k /I with I = w 1 w k i j : w i = w j. T (E) is infinite-dimensional but if dim F (E) = n then dim F (Λ E) = 2 n ; if E = v 1,..., v n then Λ k (E) = v i1 v i2 v ik i 1 < i 2 < < i k has dimension ( ) n k. Why look at exterior powers of a vector space (or its dual)?

10 (i) Exterior powers play a big role in representation theory (if you know about this). (ii) By composition, a linear map A : E F induces linear maps Λ k (E) Λ k (F ), e.g. if k = 1, a 1-tensor is always skew so Λ 1 (E) = E. If E and F have dimension n, then Λ n E and Λ n F are one-dimensional vector spaces, and Λ n A is multiplication by a scalar, which is the determinant of A. [For any tensor S E p, we get a skew-symmetric tensor Alt(S) = π Sym p sgn(π)s π, where sgn(π) ±1 and we set S π (v 1,..., v p ) = S(π(v 1 ),..., π(v p )). If S is already skew, then Alt(S) = S; elements of Λ p are exactly tensors τ E p for which τ σ = sgn(σ)τ for all σ Sym p. Now think of A as a matrix, and the usual determinant formula!] (iii) Differential forms on a manifold M are sections of Λ k T M. If we integrate a function, and change co-ordinates, the determinant of the Jacobian enters; differential forms will give us a natural way to integrate on manifolds.

11 Given a vector field X Γ(T M) on a manifold, we get an operation on functions f X f : C (M) C (M), by differentiating f in the direction of the flow of X. Precisely, integrate X to a one-parameter family of diffeomorphisms {φ t : M M} t ( δ,δ), and consider the function X f : M R with value (X f)(x) = dt d (f φ t(x)) t=0. Everything here is local, so this defines d : C (M) = Ω 0 (M) Ω 1 (M) = Γ(T M) via df, X = X f. We will extend this map to a cochain complex, d 2 = 0, Ω 0 (M) d Ω 1 (M) d Ω 2 (M) d The complex ends Ω n (M) 0 since there are no (n + 1)-forms on an n-manifold. The associated de Rham cohomology theory HdR (M) agrees with e.g. singular cohomology.

12 Tensor product on T (E) descends to give wedge product : Λ k Λ l Λ k+l. (i) Wedge product is bilinear and associative (the second of these can take some work). (ii) If φ Λ k and η Λ l then φ η = ( 1) kl η φ. [Note: such a nice formula will not hold if we start with elements of Λ (E) which are not themselves of pure degree. Also, from the defining skew-symmetry, if v E then v v = 0, but this is not true for general v Λ (E).] We can do to vector bundles anything we can do to vector spaces, so now a vector bundle E M gives vector bundles Λ k E M. If E = T M we write Ω i (M) for the space of global sections of Λ i T M. The entire algebra (Ω (M), ) is called the de Rham algebra. It s important, e.g. for smooth algebraic varieties it determines the entire Q-homotopy type [this is not obvious].

13 A smooth map f : M N induces f : Ω (N) Ω (M), by (f φ) x = (df x ) (φ f(x) ), and df x the linear map dual to df x. Each of the following is easy to check. (i) f is linear, and f (φ θ) = f φ f θ; (ii) (f g) φ = g f φ; Now work in an open set R n = U M with coordinates x i, so we have dx i Ω 1 (U) which at every point x U gives the usual basis for Tx M, dual to the basis { / x i } for T x M. Similarly, {dx I = dx i1 dx ik } i1 <...<i k give a local basis for Λ k T U. Check: (i) φ : U R dφ = i φ x i dx i ; (ii) f : U V and φ = I a I dx I f φ = I(f a I )df I, with df i = j f i y j dy j ; The first of these formulae motivates the general definition for differentiating forms.

14 Let φ = I φ I dx I Ω k (M) be a k-form, in local co-ordinates, so φ I C (U). We define a (k + 1)-form dφ as follows: dφ = I dφ I dx I where dφ I is interpreted via the usual definition of d on functions. We should check this makes sense, i.e. is invariant under changes of coordinates. Hideous direct computation works, or more efficiently: (i) In the fixed co-ordinates, d satisfies d(φ 1 + φ 2 ) = dφ 1 + dφ 2 ; d(φ 1 φ 2 ) = dφ 1 φ 2 +( 1) k φ 1 dφ 2 if φ 1 Ω k ; d 2 = 0, i.e. d(dφ) = 0 d(f φ) = f (dφ). (ii) These properties characterise d; anything satisfying these four things and agreeing with d on functions must always agree with d. The proofs of (i) and (ii) are easy exercises: note d 2 = 0 is the symmetry of mixed derivatives, and enables us to define HdR as promised.

15 Poincaré Lemma: HdR l (U) = 0 for l > 0 and U R k any star-shaped open set. [Obviously HdR 0 (U) = R comprises only constant functions.] The proof is just as in other cohomology theories: we define ι : Ω k (U) Ω k 1 (U) for which d ι + ι d = id; this forces all closed forms [elements of ker(d)] to also be exact [elements of im(d)]: dω = 0 ω = d(ιω). Suppose U is star-shaped about 0 R n, write ω = I ω I dx I and define ι(ω) x as I k j=1 ( 1) j 1 ( 1 0 tk 1 ω I (tx)dt One computes that ((d ι + ι d)ω) x is 1 I 0 kt k 1 ω I (tx) + n l=1 ) x ij dx I\{ij } t k ω I (tx)x l dt dx I x l = I ( 1 0 d dt ( t k ω I (tx)dt )) dx I = ω x.

16 L3: Special forms Differential forms in abstract generality as an algebra Ω (M) are important, but some forms are more important than others. Here we ll try and be more down to earth. Oneforms are fairly intuitive: they eat tangent vectors, and encode directional derivatives of functions. The next important class of forms are volume forms. Let E be an n-dimensional real vector space and (e 1,..., e n ), (f 1,..., f n ) two ordered bases of V. These are equivalently oriented if the matrix A : E E taking e i f i has det(a) > 0. This defines an equivalence relation on ordered bases with two equivalence classes, called orientations of E. [If dim(e) = 0, so E = {0} is a point, an orientation is just an arbitrary choice of sign ±1 labelling the point.]

17 (i) If π : E B is a rank k vector bundle, an orientation of E is a coherent choice {or b } b B of orientations for the fibres E b of E, coherent in the sense that the local trivialisations µ : π 1 (U) = U R k should take all the or b to a fixed orientation of R k. (ii) An orientation of a smooth manifold is a choice of orientation for the tangent bundle T M (if this exists, and M is connected..., there are two such). If M is oriented, write M for the same smooth manifold with the opposite orientation. (iii) An orientation of M defines canonically an orientation of its boundary M, by declaring that a basis (e 1,..., e n 1 ) of T x ( M) is positively oriented if (n x, e 1,..., e n 1 ) is a positive basis of T x M, where n x is an outward pointing normal. (iv) If M is a compact oriented one-manifold then p M or(p) = 0, where or(p) {±1} is the orientation induced at the point p.

18 Recall from calculus, if f : U V is a diffeomorphism of open subsets of R k, then V ady 1... dy k = U (a f) det(df) dx 1... dx k where a L 1 (V ) is integrable and x det(df x ) is the obvious function on U. So integration of functions is not consistently defined: V a U (a f). This is obvious: diffeomorphisms distort volume, and the determinant exactly measures how. But this means that the change-ofvariables formula exactly measures the natural transformation of k-forms on R k : φ = ady 1 dy k f φ = (φ f) det(df)dx 1 dx k If f preserves orientation, so det(df) > 0 everywhere, we accordingly have V φ = U f φ. Using a partition of unity we immediately get: Lemma: if X is an oriented k-manifold, there is a well-defined integration map X : Ωk ct (X) R, defined on the subspace of forms with compact support.

19 Note: we can only integrate k-forms over oriented k-dimensional manifolds. A volume form on a manifold M k is a nowhere zero differential form of degree k. Since Λ k T x M is only one-dimensional, a volume form picks out a distinguished isomorphism Λ k T x M R for each x, hence trivialises Λ k T M. Clearly, M admits a volume form if and only if this real line bundle is trivial, if and only if M is orientable. The choice of orientation is equivalent to a choice of volume form, up to multiplication of the latter by everywhere positive functions. Volume forms are closed, as Ω k+1 (M) = 0; the Poincaré Lemma shows they can be zero, but this can t happen on a closed manifold: Stokes Theorem: M dω = M ω. Corollary: for a closed oriented k-manifold, X : Hk dr (X) R is onto (in fact an isomorphism, but this takes a bit more work).

20 Since it s so important, we should prove Stokes theorem. Both sides are linear in the form ω, so we reduce to working in a chart U R k + open in a half-space in M, with ω compactly supported in U. [Here R k + = {x R k x 1 0}.] (i) If U R k + =, then M ω = 0 trivially, whilst M dω = U dν where ν = (incl U ω). In the local basis of forms, we write ν = k j=1 ( 1) j 1 f j dx 1... dx ˆ j... dx k, and apply Fubini s theorem to write: ( + dν = f i dx i dx 1... dx ˆ i... dx k. R k i R k 1 x i and the bracket on the RHS vanishes. (ii) If U meets R k +, we apply the above to all terms except the x 1 -integral, which goes from 0 to and gives R dν = f 1(0, x 2,..., x k )dx 2... dx k. k R k 1 But x 1 0 on R k + dx 1 0 on R k +, so ν R k + = f 1 dx 2... dx k. Now note that e 1 is the outward unit normal to finish the proof. )

21 We can go and do lots of cohomology theory with HdR : prove Poincaré duality, use Stokes theorem to study cobordism, prove H k (M) = R if M is both closed and oriented and H k (M) = 0 otherwise, where dim(m) = k. But differential forms can do more geometric things than just define cohomology classes. Let p : P B be a fibre bundle (smoothly locally trivial; if M is compact, it suffices for p to be a submersion). The fibres look the same; how can we canonically identify them? One way is to choose a connexion, that is a splitting T p P = ker(dπ) p Hor p by a field of horizontal spaces. If the fibres are 1-dimensional a circle bundle we re choosing hyperplanes in all the tangent spaces T p P. Defining a hyperplane as ker(f) for a linear functional f, we see that the connexion is defined by a one-form, θ Ω 1 (P ) with ker(θ) p T p S 1 = {0} everywhere. Such connexion forms will be important later when we talk about quotients by circle actions.

22 A two-form φ Ω 2 (M) defines a natural map µ φ : T M T M from the tangent to cotangent bundle, by setting µ φ (u), v = φ(u, v) u, v T M The 2-form is called non-degenerate if this natural map is an isomorphism. Definition: A symplectic form ω is a closed non-degenerate 2-form. A choice of such a form on a manifold, up to pullback by diffeomorphisms, is called a symplectic structure. The non-degeneracy is a pointwise condition, but being closed is a kind of integrability constraint, and is local but can t be reformulated in terms of only pointwise information. Key fact: a function determines a flow: f df (µ ω ) 1 (df) = X f, which is a vector field so can be integrated. Since this is natural, the flow preserves ω (we ll prove this more carefully later) so symplectic forms have large spaces of symmetries, not like metrics.

23 Digression: A foliated manifold is one with an atlas of charts whose transition maps are diffeomorphisms of R n = R n k R k which preserve vertical slices : φ αβ = (f(x), g(x, y)). Tangent spaces to slices define a subbundle which is integrable. If F is a codimension one (k = 1) with trivial normal bundle T M/F, there is λ Ω 1 (M) s.t. λ F = 0 but m, λ m 0. Proposition: (i) Integrability dλ = η λ for some η Ω 1. (ii) Γ = η dη is a closed 3-form (iii) The class [Γ] HdR 3 (M) is an invariant of F, so independent of choices of λ and η. This is the Godbillon-Vey class. To prove (ii), note d(η λ) = 0 dη λ = 0. This means (check!) ξ Ω 1 s.t. dη = ξ λ. Now dγ = dη dη = 0. (iii) is a similar check: for η use η η = fλ and for λ use λ = gλ, for functions f, g on M. Thurston: for any t R, there is a 2d-foliation on S 3 with [Γ] = t H 3 (S 3 ) = R.

24 L4: Lie derivative A symplectic form is a closed non-degenerate 2-form. What does it look like? Pointwise, a non-degenerate skew-form ω Λ 2 V on a vector space V is one for which (i) ω(u, v) = ω(v, u) and (ii) ω(u, v) = 0 v u = 0. Example: ω 0 = n i=1 dx i dy i on R 2n (a 2-form with constant co-efficients). Define a matrix, with respect to the basis x 1,..., x n, y 1,..., y n, ( 0 In ) J 0 =. The group Sp I n 0 2n (R ) comprises linear maps A for which A ω 0 = ω 0, i.e. ω(av, Au) = ω(v, u), which says A t J 0 A = J 0. Lemma: if V, ω is a symplectic vector space there is a basis u 1... u n, v 1... v n of V such that ω u i 0, ω v j 0 and ω(u i, v j ) = δ ij. Such a basis is called a symplectic basis.

25 Before giving the proof, note if U (V, ω), there is a symplectic orthogonal complement U ω = {v V ω(v, u) = 0 u U} U is a symplectic subspace if U U ω = {0}, which means ω U is non-degenerate. Proof of Lemma: We induct on dim(v ). By definition ω is non-degenerate, so given u 1 V, there is v V such that ω(u 1, v) 0; let v 1 = v/ ω(u 1, v) so ω(u 1, v 1 ) = 1. Then U = u 1, v 1 is a symplectic subspace of V, so (by induction) there is a symplectic basis u 2,..., u n, v 2,..., v n for U ω, and this completes the proof. Corollary: a symplectic vector space has even dimension. ω is non-degenerate if and only if ω n 0 Λ n V. The second statement is true in the standard model (R 2n, ω 0 ). If ω is degenerate, ω(u, v) = 0 v, extend u to a basis u = u 1, u 2,... u 2n of V and note ω n (u 1,..., u 2n ) = 0.

26 Corollary: A two-dimensional manifold is symplectic if and only if it is orientable. The sphere S 2k is not symplectic if k > 1. Proof: if dim(m) = 2, a non-degenerate 2- form is a volume form, and any such is obviously closed. For the second statement, if M 2k, ω is symplectic, the 2-form defines a class [ω] HdR 2 (M). The k-th cup-product of this class with itself is pointwise at T x M the nonzero form ωx k, i.e. ωk is a volume form, hence non-zero in cohomology. So [ω] k 0 HdR 2k(M) (which is just R ), so all the even cohomology groups H 2i (M) have non-zero dimension. We can easily build more examples. For instance, products of symplectic manifolds are symplectic: (M, ω) (N, σ) = (M N, ω σ). The last notation is shorthand for πm ω + π N σ. So the family Σ g1 Σ gk gives (closed) examples in all possible dimensions.

27 At the end of the last lecture, we said a function on a symplectic manifold defines a flow f df (µ ω ) 1 (df) = X f = X {φ X t } t ( δ,δ). Being a canonical construction, we expect the flow to preserve the symplectic structure. Formally, we prove such statements using the Lie derivative. If X Γ(T M) is a vector field with flow {φ t }, defined at least locally for small time, and α Ω (M) is some differential form, then L X (α) = d dt (φ t α) t=0 is the Lie derivative of α with respect to X, and measures the way α changes along the flow. This is actually a much more natural differential operator than either covariant differentiation with respect to a connexion, or exterior differentiation of forms themselves: and L X (α) is the same kind of object as α itself. We can also differentiate covariant tensors (like vector fields) as well as contravariant tensors like forms: just replace φ t by φ t.

28 The interior product ι is defined by ι X ω, u = ω, X u for X Γ(T M), u Λ p (T M), ω Ω p+1 (M). Note ι X ω = µ ω (X) in our previous notation. The following formulae are often useful: (i) L X (f) = X f for f C (M) (ii) L X (Y ) = [X, Y ] for X,Y in Γ(T M) (iii) L X (dα) = d(l X α) for α Ω (M) (iv) L X = ι X d + d ι X [Cartan s formula] (v) dθ(x, Y ) = X θ(y ) Y θ(x) θ([x, Y ]) for vector fields X,Y and θ Ω 1 (M) (i) and (iii) are immediate from the definitions. For (iv), note ι X is an antiderivation: ι X (u v) = ι X u v + ( 1) deg(u) u ι X v So L X and ι X d + dι X are both derivations preserving degree & both commute with d. Hence it suffices to check the identity on functions. This reduces to ι X (df) = df(x) = X f. We shall not prove (v). Recall [X, Y ] = 0 iff the flows generated by X and Y commute, which would certainly imply that L X (Y ) = 0.

29 To prove (ii), first drink a large glass of grappa. Then we compute, by showing the two sides take the same value on any function f, at any point m. Suppose X integrates to {φ t } and Y integrates to {ψ s }. Now (L X Y )(f) m equals d ( (dφ t Y dt φt (m) ) d t=0) f = dt (Y φ t (m) (fφ t)) t=0 Define H(t, u) = f φ t ψ u φ t (m) Define K(t, u, s) = f φ s ψ u φ t (m) both defined in a neighbourhood of the origin. Then check: (i) 2 H (t,0) = Y φt (m) (f φ t) (ii) 12 (H) (0,0) = L X Y (f) m (iii) 12 (H) (0,0) = 12 (K) (0,0,0) 23 (K) (0,0,0) (iv) 2 K (t,0,0) = (Y f) φt (m) (v) 12 (K) (0,0,0) = (X (Y f)) m, similarly: (vi) 23 (K) (0,0,0) = (Y (X f)) m So L X Y = XY Y X = [X, Y ], as required. If you re not convinced, you didn t drink enough grappa.

30 Lemma: The Hamiltonian flow of a function on a symplectic manifold preserves the symplectic form, i.e. is through symplectomorphisms. Proof: Let f : M R be smooth, with associated vector field X = (µ ω ) 1 (df), with ω the symplectic form. Cartan s formula says L X ω = ι X dω + d(ι X ω) but dω = 0 as symplectic forms are closed; and ι X ω = µ ω (X) = df so d(ι X ω) = d 2 f = 0. Hence L X ω = 0, which exactly says ω is preserved by the flow of the vector field. Example: solving Hamilton s equations H p = q; H q = ṗ is exactly following the flow of X H = ( H p, H which satisfies ι XH ω 0 = dh. Hence, classical mechanical flows are through (not necessarily linear!) symplectomorphisms of R 2n. q ),

31 L4: Complex and Kähler manifolds A special class of smooth manifolds are those which admit a complex structure ones for which the transition maps, as diffeomorphisms of R 2n, are in fact biholomorphisms after identifying R 2n = C n. Equivalently, we demand that for charts (U α, φ α ) and (U β, φ β ) we have d(φ 1 β φ α) GL n (C ). If M is a complex manifold, its (co)tangent spaces are naturally complex vector spaces, Choose real co-ordinates x i, y j on M and write T M C = C x, i y. j We define: = 1 ( i ) ; = 1 ( + i ) z j 2 x j y j z j 2 x j y j dz j = dx j + idy j ; dz j = dx j idy j Then define T M C = (T M) 1,0 (T M) 0,1, where we set C dz j = (T M) 1,0 and C dz j = (T M) 0,1.

32 We can now decompose the complex-valued differential forms on the manifold according to their (p, q)-type with respect to the above decomposition of 1-forms. Set Ω l,m (M) = { J =l, K =m α JK dz J dz K } with the α JK smooth C -valued functions. More formally, we define an almost complex structure on a manifold to be an endormorphism J : T M T M of the real tangent bundle satisfying J 2 = 1. A complex manifold has a distinguished such, locally: J( / x i ) = / y i ; J( / y j ) = / x j This extends to T M C by taking J( / z j ) = i( / z j ); J( / z j ) = i( / z j ) Then Ω 1,0 = {α α(jv) = iα(v)} and Ω 0,1 = {α α(jv) = iα(v)} are the +i, respectively i, eigenspaces for J acting on Γ(T M C ), and Ω l,m = Γ(Λ l Ω 1,0 Λ m Ω 0,1 ).

33 Exterior derivative d : C (M, C ) Ω 1 (M, C ) splits in the decomposition Ω 1 C = Ω1,0 Ω 0,1 to give d = +. Lemma: f is holomorphic if and only if f = 0. More significantly, the splitting d = + persists to all degrees; d(ω p,q ) Ω p+1,q Ω p,q+1 There was no a priori reason for the Ω p+2,q 1 component to vanish: it holds because of the holomorphicity of the co-ordinates z j when we write a form as J,K a JK dz J d z K and perform the differentiation in local co-ordinates. Corollary: d 2 = 0 2 = 0, which leads to Dolbeault cohomology : the groups H,j (M) are the cohomology groups of the complex Ω,j 1 Ω,j Ω,j+1 These are complex, and not smooth (letalone homotopy), invariants of the manifold.

34 We won t develop Dolbeault theory, but for a special class of manifolds something remarkable happens. In general, Ω k (M, C ) = l+m=k Ω l,m but this doesn t imply anything about cohomology: why should a closed form be holomorphic? A Kähler manifold is one equipped with a (1, 1)-form, i.e. a 2-form of type (1, 1) in the splitting above, for which ω = 0 = ω, and which is locally given by 2 i hjk dz j d z k, where (h jk ) is positive definite. Theorem (Hodge): If X is Kähler, then there is an isomorphism HdR k (X) = i+j=k H i,j (X). Moreover, H i,j = H j,i are naturally conjugate subspaces, in particular of equal dimension. Hence if X is Kähler, any odd Betti number b odd (X) is necessarily even. Theorem (Lefschetz): for Kähler X, wedge product [ω] k : HdR n k (X) Hn+k dr (X) is an isomorphism.

35 Note: Kähler manifolds are symplectic! For, pointwise, the closed (1, 1)-form ω satisfies ( ) ω n i n = n! det(h jk ) (dz 1 dz 1 dz n dz n ) 2 where n = dim C (M). This is the most important class of closed symplectic manifolds. Where do we find Kähler manifolds? Say ρ : M R is (strictly) plurisubharmonic if the matrix ( 2 ) ρ z j z k is positive definite everywhere. In this case, 2 i ρ is a Kähler form on M. [Note M cannot be closed why?] Example: ρ = z 2 defines ω 0 on C n = R 2n. By contrast, ρ = log( z 2 + 1) defines a Kähler form Ω on C n of finite total volume. To check the function is plush, check it at (1, 0,..., 0) ω = i ( dzj dz j ( zj dz j )( ) z j dz j ) z j z j (1 + z j z j ) 2 and note Ω is invariant under the U(n) action on C n which is transitive on the set of directions S 2n 1.

36 Theorem: any complex projective subvariety X C P n is Kähler. Proof: Let U i P n be {[z 0,..., z n ] : z i 0}, which is isomorphic to C n via ( z0 [z 0,..., z n ],..., z i 1, z i+1,..., z ) n z i z i z i z i The transition map for 2 such charts is ( 1 φ : (u 1,..., u n ), u 2,..., u ) n u 1 u 1 u 1 which takes Ω = 2 i (log z 2 + 1) to { ( i 2 log( z ) + log z 1 2 )} but since (log z 1 ) = 0 = (log z 1 ) (these are (anti)-holomorphic functions), this means φ Ω = Ω and hence the local Kähler forms on the U i patch to give a global Kähler form. Hence P n is Kähler. Now write X = {z 1 = 0,..., z m = 0}, and note/check that ρ X is still plush on the open sets U i X.

37 The various classes of manifolds in play here are all distinct. Example: C 2 \{0}/(z 2z) is a complex manifold: the quotient of a complex space by a discrete holomorphic group action. Topologically, this is S 1 S 3, which has b 2 = 0 (trivial second cohomology group). So this complex manifold cannot be symplectic (or Kähler). Example: a projective algebraic variety always has lots of subvarieties take intersections with projective hyperplanes. But some Kähler surfaces contain no complex curves if no Kähler (1, 1)-form lies inside H 2 (X; Z) H 2 dr (X) the Kähler manifold cannot be projective. Deformations of degree 4 surfaces in C P 3 (called K3 surfaces) are like this.

38 L6: Almost complex structures To study general symplectic manifolds, rather than Kähler manifolds, it is helpful to extract the homotopy-theoretic essence of having a complex structure. An almost complex structure is a bundle automorphism J : T M T M with J 2 = I. If M is a complex manifold, with complex co-ordinates z j = x j + iy j, then there is a distinguished J defined by J ( x ) = y ; J ( y ) = x. The tangent spaces to M are naturally complex vector spaces, which carry multiplication by i. More generally, for almost complex (M, J), J extends complex-linearly to T M C, and splits this space into ±i-eigenspaces T M C = T 1,0 (M) T 0,1 (M). So T 1,0 (M) = R T M, and in the complex case can also be viewed as T 1,0 (M) = C z, with j z = j x i j y. If J j comes from a system of complex co-ordinates, we say it is integrable.

39 Definition: An a.c.s. J on (M, ω) is compatible with a symplectic form ω if ω(jv, Ju) = ω(u, v); ω(v, Jv) > 0 v 0 The symmetric bilinear form g(u, v) = ω(u, Jv) defines a Riemannian metric in this case. Proposition: (M, ω) admits a compatible J. Proof: First we show this is true for a symplectic vector space V. Choose some metric g on V, and define A by ω(u, v) = g(au, v). Then A = A so AA is symmetric positive definite: g(aa v, v) = g(a v, A v) > 0 v 0 Thus AA = B.diag(λ 1,..., λ 2n ).B 1 is diagonalisable, with λ i > 0, and AA exists and equals B.diag( λ 1,..., λ 2n ).B 1. Set: J = ( AA ) 1 A JJ = I and J = J Thus J 2 = I is an a.c.s. and ω(ju, Jv) = g(aju, Jv) = g(jau, Jv) = g(au, v) ω(u, Ju) = g( JAu, u) = g( AA u, u) > 0 so J is compatible. Use this construction on each T x M, since it s canonical it works globally.

40 The proof actually shows that the space of compatible J is the same as the space of metrics, which is convex, hence contractible. Thus introducing a compatible J involves essentially no choice. Definition: a symplectic vector bundle is a vector bundle π : E B such that each fibre E b has a linear symplectic form ω b, these form a smooth global section Ω of Λ 2 E, and locally in B things are trivial: (π 1 (U), Ω) = (U R 2n, ω 0 ) Corollary: such an E canonically admits the structure of a complex vector bundle. Hence, it has a first Chern class. This is a characteristic class ; we assign to each complex vector bundle E B an element c 1 (E) H 2 (B; Z) such that c 1 (f E) = f c 1 (E) is natural under continuous maps and pullback of vector bundles. Note that choices of a.c.s J on E are all homotopic, so c 1 (E) does not change as it lives in the integral cohomology which is discrete. If E = T M we write c 1 (M) for c 1 (E).

41 There are many definitions of c 1 (E): (i) The determinant line bundle Λ r E B is pulled back by a classifying map φ E : B BU(1) = C P, i.e. Λ r E = φ E L taut, and we know H (C P ; Z) = Z[c univ 1 ] is generated by an element of degree 2; set c 1 (E) = φ E (cuniv 1 ). (ii) Take a generic section s : B Λ r E which is transverse to the zero-section, which implies its zero-set Z(s) is a smooth submanifold of B of real codimension 2. Set c 1 (E) = PD[Z(s)]. (iii) Pick a connexion d A on E, given by a matrix of one-forms (θ ij ) w.r.t. a basis s j Γ(E) of local sections of E: d A : Γ(E) Γ(E T B); s j i s i θ ij The curvature F A = d A d A, locally dθ + θ θ, is a matrix-valued 2-form in Ω 2 (End(E)), so T r(f A ) Ω 2 (B) is a 2-form. The Bianchi identity says this is closed, so defines a cohomology class, and c 1 (E) = [T r(f A )/2iπ]. One can check this class is independent of choice of connexion d A.

42 We noted c 1 (E) H 2 (M; Z) is an integral class, so does not change as we vary J E continuously. Here are some other properties: (i) c 1 (E ) = c 1 (E); (ii) c 1 (E E ) = c 1 (E) c 1 (E ); given any short exact sequence 0 E E E 0, the same formula holds; (iii) c 1 (E F ) = rk(e).c 1 (F ) + rk(f ).c 1 (E); in particular for a line bundle c 1 (L k ) = kc 1 (L). (iv) If M is compact, there is an isomorphism {C vector bundles/m} c 1 H 2 (M; Z) Isomorphism If M is complex, this can fail holomorphically. Examples: (i) c 1 (Σ g ) = 2 2g. This is Gauss-Bonnet: on a surface, both c 1 and the Euler characteristic count (signed) zeroes of a generic vector field. (ii) c 1 (L taut C P n ) = [H] H 2 (C P n ) = Z H, where [H] = PD[C P n 1 ] is the hyperplane class; c 1 (C P n ) = n + 1, by considering poles of a holo c n-form (dz 0 dz n )/(z 0... z n ).

43 Adjunction formula: if C (X 4, J) is an almost complex curve in an almost complex surface, i.e. if J preserves T C T X, then 2g(C) 2 = c 1 (X) [C] + [C] 2 So the genus of an almost complex curve is determined by its homology class. Proof: We have a SES of complex v.bundles 0 T C T X C ν C/X 0 Now deduce c 1 (T X C ) = c 1 (T C) c 1 (ν C/X ) and evaluate the terms. A small (smooth, generic) displacement of C in X gives a section of ν C/X with exactly [C] 2 zeroes, to sign. A four-manifold X has a (non-degenerate, symmetric) intersection form on H 2 (X; R ). Let b + and b denote the number of positive and negative eigenvalues, so the signature σ(x) is the difference b + b, whilst b + +b = b 2 (X). Now give X an a.c.s J, defining c 1 = c 1 (T X, J). Signature theorem (Hirzebruch): c 2 1 = 2e(X) + 3σ(X) is a topological invariant.

44 Despite this, there are lots of possible c 1 s. Say M 4 is even if every class a H 2 (M) has even square. Algebraic topology shows (i) if M is even and H 2 (M; Z) has no 2-torsion, any h s.t. h 2 = 2e + 3σ and h 0 H 2 (M; Z 2 ) is c 1 of some almost complex structure, and (ii) for any a.c.s. c 2 1 σ (8). Corollary: if X 4 admits an a.c.s. then the selfconnect sum X X does not, X X X does, etc. Proof: The condition c 2 1 σ(8) & the signature theorem show 1 b 1 + b + must be even. But b 1 (X X) = 2b 1 (X), b + (X X) = 2b + (X) etc, so for X X we have 1 b 1 + b + is odd. S 4 was not symplectic since H 2 (S 4 ) = 0. Now we see C P 2 C P 2 is not symplectic since it has no a.c.s. The proof that C P 3 C P 3 C P 3 is not symplectic takes most of gauge theory and a few hundred pages. In dimension 2n > 4, there is no known example of a triple (X, [ω], J) with [ω] H 2 (X) s.t. [ω] n 0 and J an a.c.s. and X known to admit no symplectic structure.

45 L7: Darboux theorem Remember if V is a symplectic vector space and U V is any subspace, we have a symplectic orthogonal complement U V. (i) U U = {0} U is symplectic; ω U > 0 ; (ii) U = U U is Lagrangian; ω U 0. (iii) If dim(u) = 1 necessarily U U ; (iv) If codim(u) = 1 necessarily U U. Suppose Σ (M, ω) is a hypersurface. Then T Σ defines a line subbundle of T Σ, which is integrable (the condition [X, Y ] = 0 for X, Y F is trivial for rank one distributions F T Σ). So Σ has a distinguished line field. Lemma: if Σ = φ 1 (0) then the flow of the Hamiltonian field X φ is along the direction of this line field. Proof: v T Σ, dφ(v) = 0, since φ Σ = 0. But dφ = ι Xφ ω, so this says X φ T Σ, as required.

46 Corollary: For M connected, Symp ct (M, ω) acts transitively on points. Proof: Fix x y M and a path γ from x to y with nonzero tangent vector field γ. Let U t = γ(t) γ(t) be the orthogonal complement, and choose a hypersurface Σ M near γ for which T γ(t) Σ = U t for each t [0, 1]. Define f : N(γ) R, a smooth function defined on a nhood N(γ) of γ M, to be the distance to Σ in an arbitrary Riemannian metric, so Σ = f 1 (0) near γ; then extend f to a compactly supported function on M. By the previous Lemma, the flow of X f in N(γ) is tangent to γ, hence integrating the flow defines a (compactly supported) symplectomorphism taking x to y along γ. This says a symplectic manifold looks the same everywhere, but doesn t give a local model for what that same is in principle every symplectic manifold could have different local structure. The Darboux theorem gives the actual local model.

47 Theorem (Darboux): Symplectic manifolds are locally standard. If p (M, ω) there s a chart f : D(ε) M with f(0) = p s.t. f ω M = ω 0. Remark: this shows the closed non-degenerate 2-form definition of (M, ω) is the same as the atlas of charts with transition matrices having derivatives in Sp 2n (R ) definition, from L1. Moser s proof of this is cunning. Key idea: If d dt ω t = dσ t ψ t Diff(M) with ψ t ω t = ω 0 Why is this roughly true? Well if it s true, we could differentiate the diffeomorphisms and say d dt ψ t = X t ψ t for some vector fields X t. In this case, 0 = d ( ) dt ψ t ω t = ψt dωt dt + ι X t dω t + d(ι Xt ω t ) which is just ψt d(σ t + ι Xt ω t ). But the equation σ t +ι Xt ω t = 0 uniquely defines a vector field X t. So if the flow of this vector field is defined, it gives the ψ t we were after for the key idea.

48 Corollary of argument: (Moser stability) If we have {ω t } a family of symplectic forms on a closed manifold M with [ω t ] HdR 2 (M) constant, there is f : M M with f ω 1 = ω 0. Proof: the cohomology assumption says dω t dt = dσ t are all exact forms. On a closed M, we can integrate any vector fields for all time. Proof of Darboux: fix g : D M, so g ω M and ω 0 are both symplectic forms on a disc. Composing g with an element of Sp 2n (R ), we assume these agree at 0 D. For t [0, 1], let ω t = (1 t)ω 0 + tg ω M, a family of closed forms, symplectic in a sufficiently small nhood U D of 0. Since HdR 2 (D) = 0, there is σ with dω t dt = dσ, and we can suppose σ vanishes at 0; then so does the vector field X t with ι Xt ω t = σ. Since X t vanishes at the origin, for some ε > 0 and every p D(ε) D, the trajectory of p under the flow of X t stays inside U for all t [0, 1]; so on D(ε) we can integrate up and define {ψ t } with ψ 1 g ω M = ω 0 on D(ε).

49 Remark: in Moser-type theorems, it s crucial that the cohomology class does not vary. For example, consider M = C P 1 C P 1 with the Kähler forms Ω 1 = ω ω and Ω 2 = ω tω with t > 1. Gromov showed: (i) Symp(M, Ω 1 ) SO(3) SO(3); (ii) π 1 (Symp(M, Ω 2 )) is infinite. Here the symplectic topology changes drastically with an arbitrarily small perturbation of the symplectic form. McDuff: on the blow-up of S 2 T 2 S 2 S 2 along S 2 T 2 {(pt, pt)}, there are two cohomologous symplectic forms which are not diffeomorphic there s no interpolating {ω t }. Remark: even on manifolds with HdR 2 = 0, there can be interestingly different symplectic structures. For instance, there are exotic symplectic structures on R 2k for k 2, meaning symplectic structures with no embedding into (R 2k, ω st ), in particular not globally symplectomorphic to the standard structure.

50 The Darboux theorem gives the local model for (M, ω) near a point p. There are also neighbourhood theorems near entire symplectic submanifolds Q M. First a Poincaré-type lemma, which we leave as an exercise: Lemma: Let Q M be a closed smooth submanifold, and α 1, α 2 Ω k (M) be differential forms agreeing on T M Q. Then there is a form β Ω k 1 (U Q ) defined in an open nhood of Q and with dβ = α 1 α 2, and β 0 on T M Q. Note: we assume more than i Q α 1 = i Q α 2. Theorem: If Q (M, ω) is a symplectic submanifold, the symplectic structure near Q is determined by ω Q and the normal bundle ν Q/M. Note that the normal bundle of a symplectic submanifold is canonically the symplectic orthogonal complement ν Q/M = T Q T M. A metric defines an exponential map to a tubular neighbourhood exp : ν Q/M U Q as usual.

51 Proof: Precisely, suppose Q i M i are symplectic and we have φ : Q 1, ω 1 Q1 Q 2, ω 2 Q2 a symplectomorphism, and Φ : ν Q1 /M 1 ν Q2 /M 2 an isomorphism of symplectic normal bundles. Using metrics, we get a diffeomorphism φ = exp M2 Φ exp 1 M 1 : U Q1 U Q2 such that d φ = Φ on ν Q1 /M 1. Pulling back, we get ω 1 and φ ω 2, two symplectic forms on U Q1, agreeing on T M 1 Q1. By the Poincaré Lemma, we get a 1-form dσ on some smaller U Q 1, vanishing on T M 1 Q1, with dσ = ω 1 φ ω 2 on U Q 1. We can now apply Moser s method to the path ω t = tω 1 +(1 t) φ ω 2. Arguing as for Darboux, we can integrate the vector fields to time 1 in a sufficiently small neighbourhood of Q 1, and this builds the required symplectomorphism. Corollary: a nhood of a symplectic surface C (M 4, ω) is determined to symplectomorphism by g(c), the area C ω and the value [C]2 Z.

52 L8: Weinstein s neighbourhood theorem A submanifold L (M, ω) is called Lagrangian if ω L = 0. In many ways these are the most important objects in symplectic geometry. Example: a circle S 1 Σ in a surface is trivially Lagrangian. We call a product (S 1 ) n (C n, ω 0 ) a Clifford torus. By the Darboux theorem, such Lagrangian submanifolds exist in any symplectic manifold, in the trivial homology class (contrast symplectic submanifolds!). Consider R 2n = T R n with co-ordinates q j on R n and p j = dq j on the cotangent fibres. There is a one-form λ = p j dq j Ω 1 (T R n ); a diffeomorphism q q of R n induces a diffeomorphism of T R n which preserves this one-form: q j q k dq j = dq i ; p j i q = i q k j so p j dq j p j dq j. Hence for any manifold M there is a canonical λ can Ω 1 (T M) with dλ can Ω 2 (T M) a symplectic form. p k

53 Philosophical spinach: cotangent bundles and smooth algebraic varieties are the most important examples of symplectic manifolds, and most phenomena can be understood/motivated by thinking of one or the other. We can describe the canonical one-form on T M invariantly too: at a point (q, p) T M it takes the shape λ (q,p) : v p, dπ(v), where π : T M M is projection and v T (q,p) T M. (But to see dλ is symplectic, it s still easiest to write it out in local co-ordinates.) It has lots of nice properties: Lemma: If σ : M T M is a one-form, then σ λ can = σ Ω 1 (M); this property characterises λ can. Proof: Note λ m,φ : T m,φ T M R is φ dπ m,φ. If σ(m) = (m, σ m ) T M, at u T m M, (σ λ) m (u) = σ m dπ m,σm (dσ(u)) But dπ dσ = id, so (σ λ) m (u) = σ m (u).

54 Example: the zero-section M (T M, dλ can ) is a Lagrangian submanifold. (So are the fibres Tx M of the cotangent bundle.) The first example is universal, in the following sense. Theorem (Weinstein): If L (M, ω) is Lagrangian, then a nhood U(L) M of L in M is symplectomorphic to a nhood U (L) T L of the zero-section in T L. Proof: The proof is an application of Moser s method. Choose J on T M, compatible with ω, with metric g = ω(, J ). The g-orthogonal complement to T q L T q M is JT q L (Lagrangians are totally real ). Let Φ : T M T M be the isomorphism induced by g, so Φ(f) is defined by g(φ(f), v) = f(v). Then define: φ : T L M; (q, f) exp q (JΦ q (f)) Then from defining property of exp of a metric: dφ (q,0) (v, f) = v +JΦ q (f); (v, f) T q L (T q L) where we identify T q,0 (T L) = T q L (T q L).

55 Now: (φ ω M ) (q,0) ( (v, f), (v, f ) ) = ω M q ( v + JΦf, v + JΦf ) = ω M q (v, JΦf ) ω M (v, JΦf) = g q (v, Φf ) g q (v, Φf) = f (v) f(v ) = (dλ can ) (q,0) ( (v, f), (v, f ) ) So φ : T L M is s.t. dλ can and φ ω M agree over the zero-section, i.e. on T (T L) L. The Poincaré Lemma now gives us σ Ω 1 (T L) for which dλ can φ ω M = dσ, and we can apply Moser s theorem to the family of forms ω t = tdλ can + (1 t)φ ω M as usual. Corollary: a neighbourhood of a Lagrangian submanifold L M depends symplectically only on the smooth differential topology of L itself. Example: if L M is Lagrangian, it has selfintersection [L] 2 = χ(l). If L M 4 is a homologically trivial Lagrangian surface, then L is a torus or Klein bottle. If L is an oriented Lagrangian surface, [L] 2 = 2g(L) 2.

56 Two of the most important sources of Lagrangian submanifolds are the following: Proposition (i): if f : M M is a diffeomorphism, then f ω = ω, i.e. f is a symplectomorphism, if and only if Γ f (M M, ω ω) is Lagrangian. Here Γ f denotes the graph of f. Proposition (ii): if f : M M is antisymplectic, f ω = ω, then F ix(f) is Lagrangian (where it is smooth). j dz j dz j ]; hence The proofs are trivial. Example: the antidiagonal Γ id is Lagrangian (NB though obviously the diagonal M M with the usual product symplectic form is a symplectic submanifold). Example: complex conjugation on C P n is antisymplectic [recall ω F S = 2 i the real points X(R ) X(C ) of an algebraic variety X defined over R form a Lagrangian submanifold of its complex points.

57 Corollary: a neighbourhood of id Symp(M) is homeomorphic to a nhood of 0 in the space of closed 1-forms on M. In particular, Symp(M) is locally path-connected. Proof: if f is close to id then Γ f M M is close to the (Lagrangian) antidiagonal. By Weinstein s theorem we think of Γ f U for U T M a nhood of the zero-section, and Γ f becomes the graph of some 1-form σ on M. The graph of a 1-form is Lagrangian iff the 1-form is closed: σ dλ can = dσ λ can = dσ. Corollary: Suppose (M, ω) is compact and that H 1 dr (M) = 0. Every f Symp(M) C1 -close to the identity has at least 2 fixed points. Proof: if f is C 1 -close to the identity, Γ f T M is close to the zero-section Z, and can be written as the graph of a 1-form σ. But σ = dh as HdR 1 (M) = 0, and Γ f Z p dh p = 0 p Crit(h). Now M is compact, so h has at least one maximum and one minimum.

58 Remarks: (i) The last results fails if HdR 1 (M) 0; think of translation on the torus T 2 = R 2 /Z 2. (ii) A fixed point p Fix(f) is non-degenerate if df p has no eigenvalue equal to 1 (so this is generic). Morse theory and the previous argument show: f C 1 -close to id f has at least bj (M) fixed points. The Arnol d conjecture was that one could drop the assumption that f was C 1 -close to id, finally proved (circa 2000) by Fukaya-Ono, Li-Tian, Hofer-Salamon... (iii) If HdR 1 (M) 0, the same arguments apply to Hamiltonian symplectomorphisms, i.e. those which you get by flowing along X Ht for some (perhaps time-dependent) Hamiltonian function H t : M R. (iv) Fixed points are interpreted here as intersections between Γ f and, two Lagrangian submanifolds; Floer cohomology is a general theory built on the principle it s hard to displace a Lagrangian submanifold off itself by a Hamiltonian symplectomorphism.

59 L9: Blowing up The Darboux theorem lets us implant local surgeries for instance defined using the algebraic geometry of C n into any symplectic manifold; but the symplectic set-up often still retains some quirky features. A case in point is blowing up, a surgery best performed away from airport check-in desks. Definition: Z = {(z, l) C n P n 1 z l} is the blow-up of 0 C n. It has natural maps P n 1 p Z π C n with π 1-1 away from 0 but π 1 (0) = P n 1 the space of all lines at 0 C n. The map p : Z P n 1 is the total space of the complex line bundle with c 1, [P 1 ] = 1. The restriction of ω C n ω P n 1 to Z is Kähler. If we choose local C -co-ordinates p 1,..., p n at a point p of (X 2n, ω) we can form the blow-up X of X at p differentiably, i.e. as C manifold. Theorem: The blow-up X of X is symplectic.

60 Proof: Fix a chart φ : C n X s.t. 0 p with φ ω X = ω 0 (a Darboux chart). Let σ Ω 2 ( X) be closed and (i) Poincaré dual to E = P n 1 (the exceptional divisor ), and (ii) supported in the pullback U of the chart φ to X. Since H 2 dr (U) = R, we have [σ] = 1 π [ω Z], as these both give a line in E area 1. (Note E [P 1 ] = 1 as ν E/U = O( 1) is the tautological line bundle, of Chern class -1; and remember our Kähler form on P n gave a line P 1 P n area π.) Write σ = 1 π ω Z + dψ. Using cut-off functions, choose a 1-form ψ on X s.t. ψ U = ψ; then σ = σ d ψ is s.t. σ U is exactly 1 π ω Z, and still Poincaré dual to E. Set ω X = π ω X. Claim: for ε 1 sufficiently small, ω X + ε σ is globally symplectic on X. Prove this in three separate regions. (i) On X\U true by openness of non-degeneracy. (ii) Near E true since ω X E = 0 and σ is Kähler on E, hence on some nhood. (iii) So consider a compact shell between these.

61 Look at φd(δ)\ φd( 2 δ ). In this annular shell, ω X = π ω X is exactly π ω C n, and σ = π 1 ω Z is a Kähler form. These forms are compatible with the same (integrable) complex structure J = i. But this means their convex combinations are symplectic. Remark: The final trick is incredibly useful: if for some J both ω 1 and ω 2 satisfy ω i (u, Ju) > 0 whenever u 0, the convex combinations tω 1 + (1 t)ω 2 are symplectic for all t. Examples: blow up P 2 at a point p. The complex lines through p all become disjoint in X = F 1, the first Hirzebruch surface. The exceptional divisor E = P 1 is a 2-sphere of selfintersection 1 in F 1. Each complex line in P 2 lifts to a curve hitting E once; this structure shows F 1 P 1 is fibred by lines, and E is a section of the fibration. Precisely, the map P 2 \{[1, 0, 0]} P 1, [x, y, z] [y, z] extends to the blow-up F 1. Differentiably F 1 is a nontrivial S 2 -bundle over S 2 (F 1 is not even ).

62 Remark: Smale proved Diff + (S 2 ) SO(3), hence it has π 1 = Z 2. So [fill in details!] there are only two non-trivial S 2 -bundles over S 2 ; we ve shown they re both symplectic. Example: the singular variety {z 1 z 2 = 0} P 2 lifts in F 1 to a smooth disjoint union of 2 lines. Example: Take 2 smooth cubics f, g in P 2. Generically they intersect in 9 points e.g: {x 3 + y 3 + z 3 = 0}; {x 3 + y 3 + z xyz = 0} Near the 9 points the cubics {λf + µg} [λ,µ] P 1 all vanish, and this pencil of cubics looks like the family of lines in C 2 through 0. Blow up 9 times and the cubics become disjoint, so the result E(1) P 1 is again fibred (the rational [λ,µ] elliptic surface). Note some of the cubics are singular, e.g. [ 1, 1] {xyz = 0}. If we make one singular fibre {x 3 = y 2 z} = homeo S 2 then: Corollary: there is a symplectic four-manifold X containing a symplectic T 2 of square zero with π 1 (X\T 2 ) = 0. [We will use this later on!]

63 Blowing up holomorphically is natural : every biholomorphism C n f C n lifts to one Z f Z: { (z, l) f } { (f(z), [f(z)]) ; (0, l) f (0, df0 (l)) z 0 So the complex blow-up is well-defined as a complex manifold. Symplectically things are more subtle. Define a closed 2-form on Z by ω λ = π ω C n + λ 2 p ω P n 1 Lemma: for λ > 0, ω λ is Kähler and for δ > 0, and Z(δ) = {(z, l) Z z δ}, (Z(δ)\Z(0), ω λ ) =symp ( ) B( λ 2 + δ 2 )\B(λ), ω st Given the Lemma, we define the weight λ blowup of (M, ω) by symplectically embedding a standard ball φ : B((λ 2 + δ 2 ) 1 2) M for some δ > 0, and setting M = (M\im φ) Z(δ); ω λ = ω M\im φ ω λ The Lemma shows the forms agree on the overlap. NB: vol( M, ω λ ) = vol(m, ω) vol(b(λ)) so volume decreases under blowing up. }