Hodge Theory
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Hodge Theory Victor Guillemin Notes, Spring 1997 Contents 1 Deﬁnition and properties of ?. 1 2 Exterior and interior multiplication. 3 3 The case of B symmetric positive deﬁnite. 5 4 The case of B symplectic. 6 5 Graded sl(2). 7 6 Hermitian vector spaces. 10 7 Symplectic Hodge theory. 13 8 Excursus on sl(2) modules. 15 9 The strong Lefschetz property. 19 10 Riemannian Hodge theory. 22 11 Kaehler Hodge theory. 23 1 Deﬁnition and properties of ?. Let V be an n-dimensional vector space over R and B a bilinear form on V . B induces a bilinear form on ^pV , also denoted by B determined by its value on decomposable elements as B(µ, ν) := det(B(ui; vj )); µ = u1 ^ · · · up; ν = v1 ^ · · · vp: Suppose we also have ﬁxed an element Ω 2 ^nV which identiﬁes ^nV with R. Exterior multiplication then identiﬁes ^n−pV with (^pV )∗ and B maps ^pV ! (^pV )∗. We thus get a composite map ? : ^pV ! ^n−pV 1 characterized by α ^ ?β = B(α, β)Ω: (1) Properties of ?. • Dependence on Ω. If Ω1 = λΩ then ?1 = λ? as follows immediately from the deﬁnition. • Dependence on B. Suppose that B1(v; w) = B(v; Jw); J 2 End V: Extend J to an element of ^V by J(v1 ^ · · · vp) := Jv1 ^ · · · ^ Jvp. Thus the extended bilinear forms are also related by B1(µ, ν) = B(µ, Jν) and hence ?1 = ? ◦ J: • Behavior under direct sums. Suppose V = V1 ⊕ V2; B = B1 ⊕ B2; Ω = Ω1 ^ Ω2 under the identiﬁcation ^V = ^V1 ⊗ ^V2: r s Then for α1; β1 2 ^ V1; α1; β2 2 ^ V2 we have B(α1 ^ α2; β1 ^ β2) = B(α1; β1)B(α2; β2) and (α1 ^α2)^?(β1 ^β2) = B(α1 ^α2; β1 ^β2)Ω = B(α1; β1)Ω1 ^B(α2; β2)Ω2 while α1 ^ ?1β1 = B(α1; β1)Ω1; α2 ^ ?2β1 = B(α2; β2)Ω2: Hence n1−r)s r s ?(!1 ^ !2) = (−1) ?1 !1 ^ ?2!2 for !1 2 ^ V1; !2 2 ^ V2: (n1−r)(n2−s) Since ?1!1 ^ ?2!2 = (−1) ? 2!2 ^ ?1!1 we can rewrite the preceding equation as (n1−r)n2 ?(!1 ^ !2) = (−1) ?2 !2 ^ ?1!1: 2 In particular, if n2 is even we get the simpler looking formula ?(!1 ^ !2) = ?2!2 ^ ?1!1: So, by induction, if V = V1 ⊕ · · · ⊕ Vm is a direct sum of even dimensional subspaces and Ω = Ω1 ^ · · · ^ Ωm then ?(!1 ^ · · · ^ !m) = !m ^ · · · ^ !1; !i 2 ^(Vi): (2) 2 Exterior and interior multiplication. Suppose that B is non-degenerate. For u 2 V we let eu : ^V ! ^V denote ∗ exterior multiplication by u. For γ 2 V we let iγ : ^V ! ^V denote interior multiplication by γ. We can also consider the transposes of these operators with respect to B: y p p−1 ev : ^ V ! ^ V; deﬁned by y p−1 p B(evα, β) = B(α, evβ); α 2 ^ V; β 2 ^ V and y p−1 p iγ : ^ V ! ^ V deﬁned by y p+1 p B(iγα, β) = B(α, iγ β); α 2 ^ V; β 2 ^ V: We claim that y p−1 −1 ev = (−1) ? ev ? (3) and y p −1 iγ = (−1) ? iγ ? (4) on ^pV . Proof of (3). For α 2 ^p−1V; β 2 ^pV we have B(ev ^ α, β)Ω = evα ^ ?β = (−1)p−1α ^ v ^ ?β p−1 −1 = (−1) α ^ ? ? ev ? β p−1 −1 = (−1) B(α, ? ev ? β)Ω: Proof of (4). Let α 2 ^p+1V; β 2 ^pV so that α ^ ?β = 0: 3 We have 0 = iγ(α ^ ?β) p−1 = (iγα) ^ ?β + (−1) α ^ iγ ? β p−1 −1 = (iγα) ^ ?β + (−1) α ^ ?(? iγ?)β so p −1 B(iγα, β)Ω = (−1) B(α, ? iγ ? β)Ω: y y There are alternative formulas for ev and iγ which are useful, and involve dualities between V and V ∗ induced by B. We let h ; i denote the pairing of V and V ∗, so hv; `i denotes the value of the linear function, ` 2 V ∗ on v 2 V . Deﬁne the maps op op ∗ L = LB; and L = LB : V ! V by hv; Lwi = B(v; w); hv; Lopwi = B(w; v); v; w 2 V: (5) We claim that y ev = iLopv (6) y iLv = ev (7) Proof. We may suppose that v 6= 0 and extend it to a basis v1; : : : ; vn of V , with v1 = v. Let w1; : : : ; wn be the basis of V determined by B(vi; wj ) = δij : 1 n ∗ Let γ ; : : : ; γ be the basis of V dual to w1; : : : ; wn and set γ := γ1. Then op hwi; L vi = B(v; wi) = δ1i = hwi; γi so γ = Lopv: If J = (j1; : : : ; jp) and K = (k1; : : : ; kp+1 are (increasing) multi-indices then J K B(evv ; w ) = 0 unless k1 = 1 and kr+1 = ir; r = 1; : : : ; p, in which case J K B(evv ; w ) = 1: The same is true for J K B(v ; iγw ): Hence y ev = iγ 4 which is the content of (6). Similarly, let w = w1 and β = L(w) so that iβvj = B(vj ; w1) = δ1j : Then K J B(iβ(v ); w ) = 0 unless k1 = 1 and kr+1 = jr; r = 1; : : : ; p in which case K J B(iβ(v ); w ) = 1 and the same holds for B(vK ; w ^ wJ ). This proves (7). Combining (3) and (6) gives −1 p−1 ? ev? = (−1) iLopv; (8) while combining (4) and (7) gives −1 p ? iLv? = (−1) ev: (9) On any vector space, independent of any choice of bilinear form we always have the identity ∗ iγew + ewiγ = hw; γi; v 2 V; γ 2 V : If γ = Lopv, then hw; γi = B(v; w) so (3) implies y y evew + ewev = B(v; w)I: (10) 3 The case of B symmetric positive deﬁnite. In this case it is usual to choose Ω such that kΩk = 1. The only choice left is then of an orientation. Suppose we have ﬁxed an orientation and so a choice of Ω. To compute ? it is enough to compute it on decomposable elements. So let U be a p-dimensional subspace of V and u1 ^· · ·^wp an orthonormal basis of U. Let W be the orthogonal complement of U and let w1; : : : ; wq be an orthonormal basis of W where q := n − p. Then u1 ^ · · · ^ up ^ w1 ^ · · · ^ wq = Ω: We claim that ?(u1 ^ · · · ^ up) = w1 ^ · · · ^ wq: We need only check that B(α, u1 ^ · · · ^ up)Ω = α ^ w1 ^ · · · ^ wq p for α 2 ^ V which are wedge products of ui and wj since u1; : : : ; up; w1; : : : ; wq form a basis of V . Now if any w's are involved in this product decomposition 5 both sides vanish. And if α =1 ^ · · · ^ up then this is the deﬁnition of the occurring in the formula. Suppose we have chosen both bases so that = +. Then ?(u1 ^ · · · ^ up) = w1 ^ · · · ^ wq while ?(w1 ^ · · · wq) = u1 ^ · · · up where is the sign of the permutation involved in moving all the w's past the u's. This sign is (−1)p(n−p). We conclude ?2 = (−1)p(n−p) on ^p V: (11) In particular ?2 = (−1)p on ^p V if n is even. (12) 4 The case of B symplectic. Suppose n = 2m and e1; : : : ; em; f1; : : : ; fm is a basis of V with B(ei; fj) = δij ; B(ei; ej ) = B(fi; fj) = 0: We take Ω := e1 ^ f1 ^ e2 ^ f2 · · · ^ em ^ fm which is clearly independent of the choice of basis with the above properties. If we let Vi denote the two dimensional space spanned by ei; fi with Bi the restriction of B to Vi and Ωi := ei ^ fi then we are in the direct sum situation and so can apply (2). So to compute ? in the symplectic situation it is enough to compute it for a two dimensional vector space with basis e; f satisfying B(e; f) = 1; e ^ f = Ω: Now B(e; e) = 0 = e ^ e; B(f; e)Ω = −Ω = f ^ e so ?e = e: Similarly ?f = f: On any vector space the \induced bilinear form" on ^0 is given by B(1; 1) = 1 so ?1 = Ω: 6 On the other hand, B(e; e) B(e; f) 0 1 B(e ^ f; e ^ f) = det = det = 1: B(f; e) B(f; f) −1 0 So ?(e ^ f) = 1: This computes ? is all cases for a two dimensional symplectic vector space. We conclude that ?2 = id (13) ﬁrst for a two dimensional symplectic vector space and then, from (2), for all symplectic vector spaces. 5 Graded sl(2). We consider the three dimensional graded Lie algebra g = g−2 ⊕ g0 ⊕ g2 where each summand is one dimensional with basis F; H; E respectively and bracket relations [H; E] = 2E; [H; F ] = −2F; [E; F ] = H: For example, g = sl(2) with 0 1 0 0 1 0 E = ; F = ; H = : 0 0 1 0 0 −1 Let V be a symplectic vector space with symplectic form, B and symplectic basis u1; : : : ; um; v1; : : : ; vm so B(ui; uj) = 0; = B(vi; vj ); B(ui; vj ) = δij : Let ! := u1 ^ v1 + · · · + um ^ vm: This element is independent of the choice of symplectic basis. (It is the image in ^2V of B under the identiﬁcation of ^2V ∗ with ^2V induced by B.) Let E(!) : ^V ! ^V denote the operation of exterior multiplication by !. So E(!) = eui evi : X Let F (!) := E(!)y 7 so y y F (!) = evi eui : X For α 2 ^pV we have, by (3), y y p−1 p−2 eveuα = (−1) (−1) ? veveu ? α = − ? eveu ? α = ?euev ? α so F (!) = ?E(!) ? : (14) 1 m 1 m ∗ Alternatively, if µ ; : : : ; µ ; ν ; : : : ; ν is the basis of V dual to u1; : : : ; vm then F (!) = iνj iµj : (15) X We now prove the Kaehler-Weil identity [E(!); F (!)]α = (p − m)α, α 2 ^pV: (16) Write E(!) = E1 + · · · + Em; Ej := euj evj and F (!) = F1 + · · · Fm; Fj = iνj iµj : Let Vj be the two dimensional space spanned by uj; vj and write α = α1 ^ · · · ^ αp; αj 2 ^Vj : Then Ei really only aﬀects the i-th factor since we are multiplying by an even element: Ei(α) = α1 ^ · · · ^ Eiαi ^ · · · ^ αp and Fi annihilates all but the i-th factor: Fi(α) = α1 ^ · · · ^ Fiαi ^ · · · ^ αp: So if i < j EiFj (α) = Fj Ei(α) = α1 ^ · · · ^ Eiαi ^ · · · ^ Fj αj ^ · · · ^ αp: In other words, [Ei; Fj ] = 0; i 6= j: So [E(!); F (!)]α = α1 ^ · · · ^ [Ei; Fi]αi ^ · · · ^ αp X Since the sum of the degrees of the αi add up to p, it is suﬃcient to prove (16) for the case of a two dimensional symplectic vector space with symplectic basis u; v .