Symplectic Geometry

Home , Codimension, Symplectic vector space

Symplectic Geometry

Lecture 1 of 16: Symplectic Linear Algebra Dominic Joyce, Oxford University Spring term 2021 These slides available at http://people.maths.ox.ac.uk/∼joyce/

1 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Plan of talk:

1 Symplectic Linear Algebra

1.1 Multilinear maps and exterior forms

1.2 Vector subspaces of symplectic vector spaces

1.3 The symplectic group

1.4 Symplectic and complex geometry

2 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry 1. Symplectic Linear Algebra 1.1. Multilinear maps and exterior forms

Let V be a ﬁnite-dimensional real vector space. Write Λk V ∗ for the vector space of k-forms on V . One way to deﬁne Λk V ∗ is as k the vector space of skew symmetric multilinear maps F : V → R satisfying

F (v1,..., vi−1, vi+1, vi , vi+2,..., vk ) = −F (v1,..., vk ), 0 0 F (αv1 + βv1, v2,..., vk ) = αF (v1, v2,..., vk ) + βF (v1, v2,..., vk ).

3 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Then Λ1V ∗ = V ∗, the dual vector space of V . There is an associative wedge product ∧ :Λk V ∗ × Λl V ∗ → Λk+l V ∗ given by 1 (F ∧ G)(v ,..., v ) = · 1 k+l (k + l)! X sign(σ)F (vσ(1),..., vσ(k))G(vσ(k+1),..., vσ(k+l)).

σ∈Sk+l

n k ∗ If V = R with coordinates (x1,..., xn) then Λ V has dimension n k and basis dxi1 ∧ dxi2 ∧ · · · ∧ dxik for 1 6 i1 < ··· < ik 6 n.

4 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

Let ω ∈ Λ2V ∗ be a 2-form. Deﬁne a linear mapω ˜ : V → V ∗ by ω˜ : u 7→ u · ω, that is,ω ˜(u) is the linear map V → R taking v 7→ ω(u, v). We call ω symplectic ifω ˜ : V → V ∗ is an isomorphism, and then we call (V , ω) a symplectic vector space. Remark A (pseudo)-Riemannian metric g is a symmetric bilinear map ∗ g : V × V → R giving an isomorphismg ˜ : V → V . There are similarities between Riemannian and symplectic geometry.

5 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Proposition 1.1 2 ∗ ω ∈ Λ V is symplectic if and only if dim V = 2n, some n > 1, and V admits a basis (u1,..., un, v1,..., vn) such that ω(ui , uj ) = ω(vi , vj ) = 0 and ω(ui , vj ) = δij , all i, j = 1,..., n.

2 ∗ Proof. Let ω ∈ Λ V , and suppose u1,..., un are a maximal linearly independent subset of V such that ω(ui , uj ) = 0 for all i, j. Deﬁne U = u ∈ V : ω(ui , u) = 0 for all i = 1,..., n . Then dim V − dim U 6 n, since U is deﬁned by n equations ω(ui , u) = 0. Also uj ∈ U for all j as ω(ui , uj ) = 0 for all i, so hu1,..., uni ⊆ U. If hu1,..., uni= 6 U then we can take un+1 ∈ U \ hu1,..., uni, contradicting u1,..., un maximal. So U = hu1,..., uni, and dim U = n, as u1,..., un are linearly independent.

6 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

Consider two cases: (A) dim V − dim U < n; and (B) dim V − dim U = n. In case (A), the equations ω(ui , u) = 0 for i = 1,..., n are ∗ dependent, soω ˜(u1),..., ω˜(un) are linearly dependent in V , and thusω ˜ is not an isomorphism, as u1,..., un are linearly independent. So ω is not symplectic. In case (B), since the equations ω(ui , u) = 0 for i = 1,..., n are independent, we can choose w1,..., wn such that ω(ui , wj ) = δij j−1 for all i, j. Set vj = wj − Σi=1ω(wi , wj )ui . Can check (u1,..., un, v1,..., vn) is a basis of V with ω(ui , uj ) = ω(vi , vj ) 1 n 1 = 0 and ω(ui , vj ) = δij , for all i, j = 1,..., n. Let (e ,..., e , f , n i i ..., f ) be the dual basis. Thenω ˜ maps ui 7→ f and vi 7→ −e , so ω˜ is an isomorphism, and ω is symplectic.

7 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Conclusion ∼ 2n All symplectic forms on V = R may be written in a standard form, with matrix 0 id n×n . − idn×n 0 Thus, symplectic forms on V all ‘look the same’, they are all conjugate under GL(V ) = GL(2n, R). Often we write this standard form in coordinates (x1,..., xn, y1, 2n Pn ..., yn) on R , with ω = i=1 dxi ∧ dyi .

8 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

Here is a useful criterion for when a 2-form is symplectic: Lemma 1.2 Let V be a vector space of dimension 2n. Then ω in Λ2V ∗ is symplectic if and only if ωn 6= 0 in Λ2nV ∗.

Proof. Suppose ω is symplectic. Then there exist coordinates (x1, Pn ..., xn, y1,..., yn) on V with ω = i=1 dxi ∧ dyi . So

n ω = n! dx1 ∧ dy1 ∧ · · · ∧ dxn ∧ dyn 6= 0.

Now suppose ω is not symplectic. Thenω ˜ : V → V ∗ is not an isomorphism, so as dim V = dim V ∗,ω ˜ is not injective, and there exists 0 6= v ∈ V with v · ω = 0. Then v · (ωn) = n(v · ω) ∧ ωn−1 = 0 in Λ2n−1V ∗, which implies ωn = 0, as v· :Λ2nV ∗ → Λ2n−1V ∗ is injective.

9 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Let (V , ω) be a symplectic vector space, and U 6 V a vector subspace. Deﬁne the symplectic orthogonal subspace Uω of U by

Uω = v ∈ V : ω(u, v) = 0 for all u ∈ U .

Then Uω = (˜ω)−1(U◦), where U◦ is the annihilator ∗ ∗ ∗ {α ∈ V : α|U ≡ 0} of U in V , andω ˜ : V → V the usual isomorphism. As dim U + dim U◦ = dim V , and dim Uω = dim U◦, we have dim U + dim Uω = dim V . Note that (Uω)ω = U.

10 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

We call a subspace U in V 2 ∗ symplectic if ω|U ∈ Λ U is a symplectic form; 2 ∗ ω isotropic if ω|U = 0 in Λ U , or equivalently, if U ⊆ U ; coisotropic if Uω is isotropic, or equivalently, if Uω ⊆ U; Lagrangian if U is isotropic and coisotropic, or equivalently, if Uω = U. Since dim U + dim Uω = dim V , we see that U isotropic implies 1 1 dim U 6 2 dim V , and U coisotropic implies dim U > 2 dim V , 1 and U Lagrangian implies dim U = 2 dim V .

11 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Let U 6 V be an isotropic subspace. Choose a basis u1,..., uk for U. Then ω(ui , uj ) = 0 for i, j 6 k. Extend this to a maximal linearly independent subset u1,..., un with ω(ui , uj ) = 0 for i, j 6 n. The proof of Proposition 1.1 shows this extends to a basis (u1,..., un, v1,..., vn) for V with ω of standard form. For U coisotropic we apply the same proof to Uω, isotropic. This gives: Corollary 1.3 Let (V , ω) be a symplectic vector space of dimension 2n, and U ⊆ V an isotropic, or coisotropic, or Lagrangian subspace. Then V admits a basis (u1,..., un, v1,..., vn) with ω(ui , uj ) = ω(vi , vj ) = 0 and ω(ui , vj ) = δij , for all i, j = 1,..., n, and (i) if U is isotropic then U = hu1,..., uk i, some 0 6 k 6 n. (ii) if U is coisotropic then U = hu1,..., un, v1,..., vk i, some 0 6 k 6 n. (iii) if U is Lagrangian then U = hu1,..., uni.

12 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry 1.3. The symplectic group

2n Let R have its usual coordinates (x1,..., xn, y1,..., yn), and set Pn ω = i=1 dxi ∧ dyi . Deﬁne the symplectic group Sp(n, R) to be the subgroup of GL(2n, R) preserving ω. 2 2n ∗ We can identify the orbit of ω in Λ (R ) : by Proposition 1.1 it is 2n the set of symplectic formsω ˆ on R , and by Lemma 1.2, this is 2 2n ∗ n the set ofω ˆ ∈ Λ (R ) withω ˆ 6= 0. Therefore ∼ 2 2n ∗ n GL(2n, R)/ Sp(n, R) = ωˆ ∈ Λ (R ) :ω ˆ 6= 0 .

2 2n ∗ As this is an open subset of Λ (R ) , taking dimensions gives

2 1 4n − dim Sp(n, R) = 2 (2n)(2n − 1),

so that dim Sp(n, R) = n(2n + 1).

13 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry Sp(n, R) is the Lie group of 2n × 2n matrices M with 0 I 0 I MT n M = n . −In 0 −In 0 It is a noncompact, semisimple Lie group of rank n. When n = 1 we have Sp(1, R) = SL(2, R) in GL(2, R). The Lie algebra sp(n, R) is the vector space of 2n × 2n matrices AB CD with BT = B, C T = C and D = −AT , for n × n matrices A, B, C, D. Corollary 1.3 implies: Corollary 1.4 The symplectic group Sp(n, R) acts transitively on: 2n (i) the set of isotropic k-planes U in R for 0 6 k 6 n. 2n (ii) the set of coisotropic k-planes U in R for n 6 k 6 2n. 2n (iii) the set of Lagrangian n-planes U in R .

14 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

2n Write Lagn for the Grassmannian of Lagrangian planes U in R . 2n It is a compact submanifold of the Grassmannian Gr(n, R ). ∼ Corollary 1.4 gives Lagn = Sp(n, R)/G, where G is the Lie subgroup of Sp(n, R) ﬁxing the subspace

2n {(x1,..., xn, 0,..., 0) : xi ∈ R} ⊂ R .

AB The Lie algebra g of G is the subspace of CD ∈ sp(n, R) with C = 0. 1 T This is of codimension 2 n(n + 1) in sp(n, R), since C = C . 1 Hence dim Lagn = 2 n(n + 1).

15 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

2n n Identify R with coordinates (x1,..., xn, y1,..., yn) with C with Pn 2 coordinates (z1,..., zn), where zj = xj + iyj . Let g = j=1 |dzj | be the Hermitian metric and

i Pn Pn ω = 2 j=1 dzj ∧ dz¯j = j=1 dxj ∧ dyj

n the Hermitian form on C . The complex structure 2n ∗ 2n 2n 2n J ∈ (R ) ⊗ R is the linear map R → R taking (z1,..., zn) 7→ (iz1,..., izn) in complex coordinates. Then

n P ∂ ∂ J = dxj ⊗ − dyj ⊗ . ∂yj ∂xj j=1

16 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry The unitary group U(n)

The Lie subgroup of GL(2n, R) preserving g, ω and J is the unitary group U(n), a compact semisimple Lie group of dimension n2. We can regard it as a group of 2n × 2n real matrices acting on 2n n R , or as a group of n × n complex matrices acting on C . Since 2n U(n) preserves the standard symplectic form ω on R we have U(n) ⊂ Sp(n, R). In fact U(n) is a maximal compact subgroup of Sp(n, R), and Sp(n, R)/ U (n) is contractible.

17 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

Going through the proof of Corollary 1.3 taking u1,..., un to be orthonormal with respect to g, we can choose u1,..., un, v1,..., vn orthonormal, so we make U, ω, g, J all of standard form. This gives: Corollary 1.5 The unitary group U(n) acts transitively on: 2n (i) the set of isotropic k-planes U in R for 0 6 k 6 n. 2n (ii) the set of coisotropic k-planes U in R for n 6 k 6 2n. 2n (iii) the set of Lagrangian n-planes U in R .

18 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Multilinear maps and exterior forms Symplectic Linear Algebra Vector subspaces of symplectic vector spaces Symplectic manifolds and Lagrangian submanifolds The symplectic group Symplectic and complex geometry

The subgroup of U(n) ﬁxing the Lagrangian plane

2n {(x1,..., xn, 0,..., 0) : xi ∈ R} ⊂ R

is the orthogonal group O(n). So ∼ Lagn = U(n)/ O (n).

+ We also write Lagn for the Grassmannian of oriented Lagrangian 2n planes in R . It is a submanifold of the Grassmannian + 2n 2n Gr (n, R ) of oriented n-planes in R , with + ∼ Lagn = U(n)/ SO(n).

It is a double cover of Lagn.

19 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra

+ Identifying Lagn with U(n)/ SO(n), and regarding U(n) and SO(n) as groups of n × n complex and real matrices, deﬁne the + Maslov map µ : Lagn → U(1) by µ : M SO(n) 7→ detC M. Now ∼ 1 1 1 R U(1) = S . Let α ∈ H (S , Z) be unique with S1 α = 1. Then ∗ 1 + µ (α) ∈ H (Lagn , Z) is the Maslov class. It is important in the algebraic topology of Lagrangians, and the theory of J-holomorphic curves.

20 / 38 Dominic Joyce, Oxford University Lecture 1: Symplectic Linear Algebra Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds

Symplectic Geometry

Lecture 2 of 16: Symplectic manifolds and Lagrangian submanifolds Dominic Joyce, Oxford University Spring term 2021 These slides available at http://people.maths.ox.ac.uk/∼joyce/

21 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds

Plan of talk:

2 Symplectic manifolds and Lagrangian submanifolds

2.1 Symplectic manifolds

2.2 Submanifolds of symplectic manifolds

2.3 (Almost) complex manifolds and K¨ahlermanifolds

22 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds 2. Symplectic manifolds and Lagrangian submanifolds 2.1. Symplectic manifolds

Let M be a smooth manifold. A symplectic form ω ∈ C ∞(Λ2T ∗M) on M is a smooth 2-form on M such that: (i) dω = 0, i.e., ω is closed; and (ii) (TpM, ω|p) is a symplectic vector space for each p ∈ M. Then we call (M, ω) a symplectic manifold. Lemma 1.2 shows that (ii) is equivalent to ωn 6= 0 at every p ∈ M, where dim M = 2n. Note that symplectic manifolds are always even dimensional. Let (M, ω) be symplectic, with dim M = 2n. Since ωn 6= 0 everywhere, there is a natural orientation on M in which ωn is a positive form. As dω = 0 we can form the de Rham cohomology 2 class [ω] in H (M, R).

23 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds

If M is compact then the natural orientation gives a fundamental n n R n class [M] ∈ H2n(M; Z), and [M] · [ω] = [M] · [ω ] = M ω > 0, n 2 as ω is a positive form. Hence the class [ω] ∈ H (M, R) satisﬁes n 2n [ω ] 6= 0 in H (M, R). These give topological obstructions to the existence of symplectic 2n forms on a manifold M. For instance, RP has no symplectic structure as it is not orientable. Also, S2 × S4 has no symplectic structure as there exists no α in 2 2 4 3 H (S × S , R) with α 6= 0, even though 2 2 4 H (S × S , R) = R 6= {0}.

24 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds

A symplectomorphism between two symplectic manifolds (M1, ω1) ∗ and (M2, ω2) is a diffeomorphism f : M1 → M2 with f (ω2) = ω1. This is the natural notion of isomorphism of symplectic manifolds. A motivating problem in symplectic geometry is to classify symplectic manifolds up to symplectomorphism. Later we will see that the group of self-symplectomorphisms of (M, ω) is infinite-dimensional. Thus, symplectic structures are not at all rigid, but very flexible.

25 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds 2.2. Submanifolds of symplectic manifolds Let (M, ω) be a symplectic manifold, and N a submanifold of M. As in §1.2, we call N: symplectic if TpN is symplectic in (TpM, ωp) for all p ∈ N, that is, (N, ω|N ) is symplectic; isotropic if TpN is isotropic in (TpM, ωp) for all p ∈ N, that is, ω|N ≡ 0; coisotropic if TpN is coisotropic in (TpM, ωp) for all p ∈ N; Lagrangian if TpN is Lagrangian in (TpM, ωp) for all p ∈ N, 1 that is, ω|N ≡ 0 and dim N = 2 dim M. We are mostly interested in Lagrangian submanifolds. If L is a Lagrangian submanifold in (M, ω) then ω|L ≡ 0, so we can form 2 the class [ω] in the relative de Rham group H (M, L, R). Also, if L is any n-dimensional submanifold of M, a necessary condition for L 0 2 to be isotopic to a Lagrangian L is that [ω|L] = 0 in H (L, R), since [ω|L] is invariant under deformations of L in M. 26 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds 2n Example: Lagrangians in R

2n Let R have coordinates (x1,..., xn, y1,..., yn), and its usual Pn symplectic form ω = i=1 dxi ∧ dyi . n Let ei : R → R be a smooth function for i = 1,..., n. Consider the submanifold L = (x1,..., xn, e1(x1,..., xn),

..., en(x1,..., xn): xi ∈ R

2n in R . When is L a Lagrangian submanifold?

27 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Well, we have

n P ω|L = dxi ∧ dei (x1,..., xn) i=1 n P ∂ei = (x1,..., xn)dxi ∧ dxj ∂xj i,j=1 n P ∂ei ∂ej = − dxi ∧ dxj . ∂xj ∂xi 16i

∂e Thus L is Lagrangian iff ∂ei ≡ j , for all i, j = 1,..., n, ∂xj ∂xi Pn n iff α = i=1 ei (x1,..., xn)dxi is a closed 1-form on R , n iff there exists smooth f : R → R with ∂f ei (x1,..., xn) ≡ (x1,..., xn). ∂xj

28 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds 2n Example: Lagrangians in R

n 2n Thus, Lagrangians close to the standard R in R are n parametrized by closed 1-forms on R , and form an inﬁnite-dimensional family. So, there are very many Lagrangians. We can also ﬁnd examples of compact Lagrangian submanifolds in 2n R . For instance, if a1,..., an > 0 then L = (x1,..., xn, y1,..., yn): 2 2 xi + yi = ai , i = 1,..., n

n 2n is a Lagrangian T in R .

29 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

2n Pn Deﬁne a 1-form θ on R by θ = − i=1 yi dxi . Then ω = dθ. If L 2n is Lagrangian in R then dθ|L = ω|L ≡ 0, so θ|L is closed, and 1 [θ|L] ∈ H (L, R) in de Rham cohomology. We call L an exact Lagrangian if θ|L is exact, that is, [θ|L] = 0 in 1 H (L, R). This is independent of the choice of θ with dθ = ω. Using J-holomorphic curve techniques (second half of lecture course), one can prove: Theorem There are no compact, embedded, exact Lagrangian submanifolds 2n in R for n > 0. 1 If H (L, R) = 0 then L is automatically exact. So, for instance, n 2n there are no embedded Lagrangian spheres S in R for n > 2.

30 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds Cotangent bundles as symplectic manifolds

Let N be a manifold, and T ∗N its cotangent bundle, with projection π : T ∗N → N. Write points of T ∗N as (p, α) with ∗ p ∈ N and α ∈ Tp N, so that π :(p, α) 7→ p. Define the ˆ ∗ ˆ ∗ tautological 1-form θ on T N by θ|(p,α) = −dπ (α), with ∗ dπ : T(p,α)(T N) → TpN the derivative of π, and ∗ ∗ ∗ ∗ ˆ dπ : Tp N → T(p,α)(T N) its dual. Defineω ˆ = dθ, a closed 2-form on T ∗N. ∗ We will showω ˆ is a symplectic form on T N. Let (x1,..., xn) be local coordinates on N. Define local coordinates (x1,..., xn, y1, ∗ ..., yn) on T N such that (x1,..., xn, y1,..., yn) represents ∗ Pn (p, α) ∈ T N if (x1,..., xn) represents p in N and α = i=1 yi dxi ∗ ˆ Pn in Tp N. Then θ = − i=1 yi dxi in local coordinates, so Pn ωˆ = i=1 dxi ∧ dyi . This is our canonical form, soω ˆ is symplectic.

31 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Note thatω ˆ is independent of choice of coordinates, and depends 2n ∗ n only on N as a manifold. Also, (R , ω) is (T N, ωˆ) when N = R . ∗ The fibres of π, the spaces Tp N for p ∈ N, are Lagrangian in T ∗N. The zero section N = {(p, 0) : p ∈ N} in T ∗N is Lagrangian in T ∗N. More generally, if α ∈ C ∞(T ∗N) is a smooth 1-form on N, then the graph of α, Γα = (p, α(p)) : p ∈ N ∗ is a submanifold of T N, and Γα is Lagrangian in N iff α is a ˆ ∗ ∗ closed 1-form, since θ|Γα = −π (α), soω ˆ|Γα = −π (dα), giving ∗ ωˆ|Γα ≡ 0 iff dα ≡ 0. Thus there are many Lagrangians in T N.

32 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds 2.3. (Almost) complex manifolds and K¨ahlermanifolds

Let M be a 2n-manifold. An almost complex structure J on M is a b b c c ∞ tensor Ja on M with Ja Jb = −δa . For v ∈ C (TM) define b b a 2 (Jv) = Ja v . Then J = −1, so J makes the tangent spaces TpM into complex vector spaces. Write [v, w] for the Lie bracket of vector fields v, w on M. a The Nijenhuis tensor N = Nbc of J satisfies a b c Nbc v w = [v, w] + J [Jv, w] + [v, Jw] − [Jv, Jw]a for all v, w ∈ C ∞(TM). We call J a complex structure if N ≡ 0, and then (M, J) is a complex manifold. If N ≡ 0 there are many holomorphic functions b locally, where f : M → C is holomorphic if Ja (df )b ≡ i(df )a, and we can make holomorphic coordinate systems.

33 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds K¨ahlermanifolds

Definition Let (M, J) be complex, and g a metric on M. We call g Hermitian c d b if gab = Ja Jb gcd . Then the Hermitian form ω is ωac = Ja gbc . It is b a 2-form on M, and gac = ωabJc . We call g Kähler and ω its Kählerform if either (i) dω = 0, (ii) ∇J = 0, or (iii) ∇ω = 0, where ∇ is the Levi-Civita connection of g, and (i)–(iii) are equivalent.

34 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds K¨ahlerpotentials

Let (M, J) be a compact complex manifold. A closed real (1,1)-form ω on M is the Kählerform of a Kählermetric g if and only if ω is positive, that is, ω(v, Jv) > 0 for all nonzero vectors v. If φ ∈ C ∞(M) then i∂∂φ¯ is a closed real (1,1)-form, and every closed real (1,1)-form is locally of this form. So locally we have ω = i∂∂φ¯ for some Kählerpotential φ : M → R. But globally we cannot solve ω = i∂∂φ¯ , as i∂∂φ¯ is exact, but ω is not. The Kählerclass of g is the de Rham cohomology class [ω] of 2 0 ω in H (M, R). Two Kählermetrics g, g lie in the same Kähler class iff ω0 = ω + i∂∂φ¯ , that is, they differ by a Kähler potential. Any small φ ∈ C ∞(M) gives a new Kählermetric g 0. So Kähler metrics on M come in infinite-dimensional families.

35 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Let (M, J) be a complex manifold, and N a submanifold of M. We call N a complex submanifold if J(TpN) = TpN in TpM for all p ∈ N. Then (N, J|N ) is a complex manifold. If g is a Kählermetric on M with Kählerform ω, then g|N is a Kählermetric on (N, JN ) with Kählerform ω|N . n Complex projective space CP with the Fubini–Study metric g is a n compact Kähler2n-manifold. Complex submanifolds N of CP are called projective complex manifolds, and they are Kählerwith metric g|N . All projective manifolds are the zero sets of finitely many polynomials in the homogeneous coordinates [z0,..., zn] on n CP . This gives huge numbers of examples of compact Kähler manifolds.

36 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds K¨ahlermanifolds and symplectic manifolds

Let (M, J) be a complex manifold with Kählermetric g, and Kählerform ω. Then dω = 0, and §1.4 implies that ω is a symplectic form at each p ∈ M, so (M, ω) is a symplectic manifold. Thus, complex projective manifolds provide huge numbers of examples of compact symplectic manifolds. Not all compact symplectic manifolds (M, ω) come from Kähler manifolds. For example, using Hodge theory one can show that the odd Betti numbers b2k+1(M) of a Kählermanifold must be even, but there exist examples of compact symplectic manifolds with b1(M) odd.

37 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds

Symplectic manifolds Symplectic Linear Algebra Submanifolds of symplectic manifolds Symplectic manifolds and Lagrangian submanifolds (Almost) complex manifolds and K¨ahlermanifolds

Let (M, ω) be a symplectic manifold, and J an almost complex structure on M. We say that J, ω are compatible if ω(v, w) = ω(Jv, Jw) and ω(v, Jv) > 0 for all nonzero vectors v, w on M. Then g(v, w) = ω(v, Jw) defines a Riemannian metric g on M. If J is an integrable complex structure, then g is Kähler,with Kählerform ω. Every symplectic manifold (M, ω) admits compatible almost complex structures J. Without assuming J is integrable, one can try to treat (M, J, ω) as a kind of Kählermanifold, and do complex geometry with it. Most of complex geometry does not generalize to this situation, but the theory of complex curves in (M, J, ω) does. We discuss this in the second half of the course.

38 / 38 Dominic Joyce, Oxford University Lecture 2: Symplectic manifolds and Lagrangian submanifolds