Symplectic Topology Example Sheet 2

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Symplectic Topology Example Sheet 2 Symplectic Topology Example Sheet 2 Dietmar Salamon ETH Z¨urich 27 February 2013 For a real vector space V denote the space of symplectic bilinear forms, linear symplectic structures, respectively inner products by S(V ) := f! : V × V ! R j ! is skew symmetric and nondegenerateg ; J (V ) := J 2 GL(V ) j J 2 = −1l ; M(V ) := fg : V × V ! R j g is an inner productg : A linear complex structure J 2 J (V ) is called compatible with the sym- plectc form ! 2 S(V ) if !(J·;J·) = !; !(v; Jv) > 0 for all v 2 V n f0g: It is called compatible with the inner product g 2 M(V ) if g(J·;J·) = g: A symplectic form ! 2 S(V ) is called compatible with the inner product g 2 M(V ) if there exists a linear complex structure J 2 J (V ) such that g = !(·;J·). For g 2 M(V ) and ! 2 S(V ) define J (V; !) := fJ 2 J (V ) j J is compatible with !g ; S(V; g) := f! 2 S(V ) j ! is compatible with gg ; J (V; g) := fJ 2 J (V ) j J is compatible with gg ; SJ (V ) := f(!; J) 2 S(V ) × J (V ) j J is compatible with !g : The first of these manifolds is contractible, the second and third are diffeo- morphic to each other, and the last three are homotopy equivalent to each other and to J (V ) and S(V ). 1 Exercise 2.1. Let V be a 2n-dimensional real vector space and let J 2 J (V ) and ! 2 S(V ). Prove that J is compatible with ! if and only if the formula gJ := !(·;J·) defines an inner product on V . Prove that, if J is compatible with !, then J is compatible with gJ . Prove that the following are equivalent. (a) J 2 J (V; !) (b) There exist v1; : : : ; vn 2 V such that the vectors v1; Jv1; : : : ; vn; Jvn form a basis of V and !(vi; Jvj) = δij and !(vi; vj) = !(Jvi; Jvj) = 0 for all i; j. 2n ∗ (c) There is a vector space isomorphism Ψ : R ! V such that Ψ ! = !0 ∗ and Ψ J = J0. Exercise 2.2. Prove that J (V; !) 6= ; for every ! 2 S(V ). Exercise 2.3. Fix an inner product g 2 M(V ). Prove that the projection πS : SJ (V ) !S(V ) is a homotopy equivalence with homotopy inverse S(V ) !SJ (V ): ! 7! (!; Jg;!): −1 Here Jpg;! := Q A 2 J (V; !) \J (V; g) is defined by g(A·; ·) = ! and Q := A∗A is the unique g-self-adjoint, g-positive-definite automorphism of V whose square is A∗A = −A2. Exercise 2.4. Fix an inner product g 2 M(V ). Prove that the projection πJ : SJ (V ) !J (V ) is a homotopy equivalence with homotopy inverse g(J; ·; ·) − g(·;J·) J (V ) !SJ (V ): J 7! (! ;J);! := : g;J g;J 2 Exercise 2.5. Prove that the map 2n −1 GL(2n; R) !J (R ):Ψ 7! ΨJΨ descends to a diffeomorphism GL(2n; R)=GL(n; C) !J (R2n). (Here we identify GL(n; C) with the subgroup of all nonsingular real 2n × 2n-matrices that commute with J0.) Exercise 2.6. Prove that the map 2n −1 ∗ GL(2n; R) !S(R ):Ψ 7! (Ψ ) !0 descends to a diffeomorphism GL(2n; R)=Sp(2n) !S(R2n). 2 Exercise 2.7. Prove that the map 2n −1 ∗ −1 GL(2n; R) !SJ (R ):Ψ 7! (Ψ ) !0; ΨJ0Ψ descends to a diffeomorphism GL(2n; R)=U(n) !SJ (R2n). 2n Exercise 2.8. Let g0 denote the standard inner product on R . Prove that 2n 2n the spaces J (R ; g0) and S(R ; g0) are both diffeomorphic to the homoge- neous space O(2n)=U(n). Construct explicit diffeomorphisms. 2n×2n 2n Exercise 2.9. Let J 2 R . Prove that J 2 J (R ;!0) if and only if the matrix P := −J0J is symplectic, symmetric, and positive definite. Deduce 2n that the space J (R ;!0) is contractible. Exercise 2.10. Denote Siegel upper half space by n×n T T Sn := Z = X + iY 2 C j X = X ;Y = Y > 0 : Prove that a matrix AB Ψ = ; A; B; C; D 2 n×n; CD R is symplectic if and only if AT D − CT B = 1l;AT C = CT A; BT D = DT B: Prove that the formula −1 Ψ∗Z := (AZ + B)(CZ + D) defines a transitive group action of Sp(2n) on Siegel upper half space. Prove ∼ that Sp(2n)\SO(2n) = U(n) is the stabilizer subgroup of the matrix i1l 2 Sn. Exercise 2.11. Prove that the formula 2n 2n −1 Sp(2n) × J (R ;!0) !J (R ;!0) : (Ψ;J) 7! ΨJΨ defines a transitive group action of the linear symplectic group Sp(2n) on 2n the space of !0-compatible linear complex structures on R , and that the ∼ stabilizer subgroup of J0 is Sp(2n) \ SO(2n) = U(n). Deduce that there is 2n a unique Sp(2n)-equivariant diffeomorphism Sn !J (R ;!0): Z 7! J(Z); such that J(i1l)= J0. Prove that an explicit formula for this diffeomorphism is given by XY −1 −Y − XY −1X J(X + iY ) = : (1) Y −1 −Y −1X 2n Deduce that J (R ;!0) is contractible. 3 Exercise 2.12. Let ! 2 S(V ). A linear complex structure J 2 J (V ) is called !-tame if !(v; Jv) > 0 for all v 2 V n f0g: This exercise shows that the space Jτ (V; !) of all !-tame linear complex structures on V is contractible. Let J 2 R2n×2n. Prove that the following assertions are equivalent. 2n (a) J 2 Jτ (R ;!0). −1 −1 (b) The matrix Z := −J0J satisfies Z = J0 ZJ0 and v; Zv > 0 for every nonzero vector v 2 R2n. −1 (c) The matrix W := (1l−Z)(1l+Z) satisfies kW k < 1 and J0W +WJ0 = 0. 2n×2n 2n The set of matrices W 2 R satisfying (c) is convex, hence Jτ (R ;!0) is contractible, and hence so is the space Jτ (V; !) of !-tame linear complex structures for every symplectic vector space (V; !). Exercise 2.13. Assume dim(V ) = 2 and let ! 2 S(V ) and J 2 J (V ). Prove that ! and J are compatible if and only if they induce the same orientation on V . Exercise 2.14. Assume dim(V ) = 4 and let g 2 M(V ). Prove that J (V; g) is diffeomorphic to two disjoint copies of the 2-sphere. Exercise 2.15. Assume dim(V ) = 6 and let g 2 M(V ). Prove that J (V; g) is diffeomorphic to two disjoint copies of a 2-sphere bundle over a 4-sphere. 4.
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