Chapter 3 Symplectic and Kaehler Geometry Lecture 12 Today: Symplectic geometry and Kaehler geometry, the linear aspects anyway. Symplectic Geometry Let V be an n dimensional vector space over R, B : V × V R a bilinea re form on V .→ Definition. B is alternating if B(v, w) = −B(w, v). Denote by Alt2(V ) the space of all alterna ting bilinea r forms on V . Definition. Take any B ∈ Alt(V ), U a subspace of V . Then we can define the orthogona l complement by U⊥ = {v ∈ V, B(u, v) = 0, ∀u ∈ U} Definition. B is non-degenerate if V⊥ = {0} . Theorem. If B is non-degenerate then dim V is even. Moeover, there exists a basis e1, . . . , en, f1, . . . , fn of V such that B(ei, en) = B(fi, fj ) = 0 and B(ei, fj ) = δij Definition. B is non-degenerate if and only if the pair (V, B) is a symplectic vector space. Then ei’s and fj ’s are called a Da rboux basis of V . Let B be non-degenerate and U a vector subspace of V Remark: dim U⊥ = 2n − dim V and we have the following 3 scenarios. 1. U isotro pic ⇔ U⊥ ⊃ U. This implies that dim U ≤ n 2. U Lagrangian ⇔ U⊥ = U. This implies dim U = n. 3. U symplectic ⇔ U⊥∩U = ∅. This implies that U⊥ is symplectic a nd BU and BU ⊥ are non-degenerate. | |Let V = Vm be a vector space over R we have Alt2(V ) ∼=Λ2(V∗) is a canonical identification. Let v1, . . . , vm be a basis of v, then Alt21 � B(vi, vj )vi ∗ ∧ vj∗(V ) ∋ B 7→ 2 and the inverse Λ2(V∗) ∋ ω 7→ Bω ∈ Alt2(V ) is given by B(v, w) = iW (iV ω) Suppo se m = 2n.6Theorem. B ∈ Alt2(V ) is non-degenerate if ωB ∈ Λ2(V ) satisfies ωn = 0B 1/2 of Proof. B non-degenerate, let e1, . . . , fn be a Darboux basis of V then ωB = � ei ∗ ∧fj∗ and we can show ωn = n!e∗ = 0B 1 ∧ f∗ n ∧f∗ n1 ∧ ··· ∧ e∗ 6Notation. ω ∈ Λ2(V ∗), symplectic geometers just say “Bω (v, w) = ω(v, w)”. Kaehler spaces V = V 2n , V a vector space over R, B ∈ Alt2(V ) is non-ge nerate. Assume we have another piece of s tructure a map J : V → V that is R-linear and J2 = −I. Definition. B and J are compatible if B(v, w) = B(Jv, Jw). Exercise(not to be handed in) Let Q(v, w) = B(v, Jw) s how that B and J are compatible if and only if Q is symmetric. From J we can make V a vector space over C by setting √−1v = Jv. So this gives V a structure of complex n-dimensional vector space. Definition. Take the bilinear form H : V × V C by →1 H(v, w) = √−1(B(v, w) + √−1Q(v, w)) B and J are compatible if and only if H is hermitian on the complex vector space V . Note that H(v, v) = Q(v, v). Definition. V, J, B is Ka hler if either H is positive definite or Q is positive definite (these two are equivalent). Consider V ∗ ⊗ C = HomR(V, C), so if l ∈ V ∗ ⊗ C then l : V C.→Definition. l ∈ (V ∗)1,0 if it is C-linear, i.e. l(Jv) = √−1l(v). And l ∈ (V ∗)0,1 if it is C-antilinear, i.e. l(Jv) = −√−1l(v). ¯Definition. lv = l(v). J∗l(v) = lJ(v). ¯Then if l ∈ (V ∗)1,0 then l ∈ (V ∗)0,1 . If l ∈ (V ∗)1,0 then J∗l = √−1l, l ∈ (V ∗)0,1 , J∗l = −√−1l. So we can decompose V ∗ ⊗ C = (V ∗)1,0 ⊕ (V ∗)0,1 i.e. decomposing into ±√−1 eig e nspace of J∗ and (V ∗)0,1 = (V ∗)0,1 . This decomposition gives a decomposition of the exterior algebra, Λr(V ∗ ⊗C) = Λr (V ∗) ⊗ C. Now, this decomposes into bigraded pieces Λr (V ∗ ⊗ C) � Λk,l(V ∗)= k+l=r Λk,l(V ∗) is the linear span of k, l forms of the form νl µiνj ∈ (V ∗)1,0 µ1 ∧ ··· ∧ µk ∧ν¯1 ∧ ··· ∧ ¯Note that J∗ : V ∗ ⊗ C → V ∗ ⊗ C can be ex tended to a map J∗ : Λr (V ∗ Λr(V ∗ ⊗ C) by setting ⊗ C) →=J∗(l1 ∧ ··· ∧ lr ) J∗l1 ∧ ··· ∧ J∗lr on decomposable elements l1 ∧ ··· ∧ lr ∈ Λr . We can define complex conjugation on Λr (V ∗ ⊗ C) on deco mposable elements ω = by ¯ ¯l1 ∧ ··· ∧ lr ¯ω = lr. Λr(V ∗ ⊗C) = Λr(V ) ⊗C, then ¯ ω ∈ Λl,k(V ∗) l1 ∧ ··· ∧ ω = ω if and only if ω ∈ Λr(V ∗) . And if ω ∈ Λk,l(V ∗) then ¯6 Proposition. O n Λk,l(V∗) we have J∗ = (√−1)k−l Id. Proof. Take ω = νl, µi, νi ∈ (V∗)1,0 then µ1 ∧ ··· ∧ µk ∧ ν¯1 ∧··· ∧ ¯J∗ω = J∗µ1 ∧ ··· ∧ J∗µk ∧ J∗ν¯1 ∧ ··· ∧ J∗ν¯l = (−1)k (−√−1)lω Notice tha t for the following decomposition of Λ2(V ⊗ C) the eigenvalues of J∗ are given below Λ2(V ⊗ C) = Λ2,0 ⊕Λ1,1 ⊕Λ0,2 � �� � ���� ���� ���� 1 −1 −1J∗ So if ω ∈ Λ∗(V∗ ⊗ C) then if Jω = ω. Now, back to serious Kahler stuff. Let V, B, J be Kahler. B 7→ ωB ∈ Λ2(V∗) ⊂ Λ2(V∗) ⊗ C. B is J invariant, so ωB is J-invariant, which happens if and only if ωB ∈ Λ1,1(V∗) and ωB is real if and ¯only if ωB = ωB . So there is a -1 correspondence between J invariant elements of Λ2(V ) and elements ω ∈ Λ1,1(V∗) which are real. Observe: (V∗)1,0 ⊗ V∗)0,1 ρ Let µ1, . . . , µn be a basis of (V∗)1,0 . Take −→ Λ1,1(V∗) by µ ⊗ ν 7→ µ ∧ ν. µj ∈ (V∗)1,0 ⊗ (V∗)0,1α = � aij µi ⊗ ¯Take ρ(α) = � aij µi ∧ ¯µj 1is it true that ρ(α) = ρ(α). No, not always. This happens if aij = −aij , equiva lently √−1[aij ] is Hermitian. We have Alt2= ωB ∈ Λ1,1(V∗)(V ) ∋ B 7→ ω 1Take α = ρ−1(ω), H = √−1 α. Then H is Hermitian. 1Check that H = √−1(B + √−1Q), B Kahler iff and only if H is positive