For the Hermitian Lefschetz operator on a complex -manifold, the Lefschetz commutator and the commutator derivation identity imply this formula. When expanding a product commutator, one must evaluate on the degree after the powers of to its right have acted.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 17 4 Solution Created 2026-10-03 Updated 2026-10-07
Let denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use . The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus , and maps bidegree to . If is the complex-linear extension of the real Hodge star operator, then . The operator is real and on degree , soUsing the complex-linear star instead would give a different bidegree, ; keeping the convention explicit prevents that ambiguity.
The Hermitian Lefschetz operator is exterior multiplication by . Define pointwise byIt lowers bidegree by . In a unitary real coframe with , it is , the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against to obtainNo integration by parts is needed for this order-zero operator. Thus is both the pointwise and the global formal adjoint.
Write , and let denote the degree of the input form. The given Lefschetz commutator is . For the required formula is exactly this identity. If it holds for , the commutator derivation identity givesThe second commutator acts on degree , so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator
For injectivity of powers of the Lefschetz operator, is immediate. Suppose and . If or , then , and impliesThe scalar is nonzero, since makes . Repeating with the smaller power shows . For , induct on , treating all cases as already established. The same commutator calculation yields, with ,The induction hypothesis applies to on degree , since and . Hence with of degree . Moreover . Since , induction also makes injective on that degree, so and . All operators preserve restriction to open sets; the argument applies on every open set. Therefore is an injective morphism of sheaves whenever .
On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since , this givesTo derive the Kähler Laplacian identity, put and , so and . Write for an anticommutator. The Kähler identities say and . Thereforebecause expansion leaves only terms containing or . Furthermore,Expanding these expressions and using makes them equal. Since and , the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies . Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with .
For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If and , then . Taking its inner product with gives . Thus is cohomologous to . Conversely, a harmonic form is -closed, and a harmonic exact form satisfies . This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.
Complex conjugation commutes with the real operator and interchanges the bidegrees and . It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with , as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic -forms bijectively to harmonic -forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives
For the final hard Lefschetz isomorphism on Dolbeault cohomology, take and . This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by , and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since . Hodge symmetry and Hodge duality give , so it is a bijection between finite-dimensional spaces of equal dimension. Because , the map on harmonic representatives agrees with exterior multiplication by on Dolbeault cohomology. We concludeFor , the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree to ; a negative exterior-multiplication power is not defined.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 4 c Solution Created 2026-10-03 Updated 2026-10-06
A positive holomorphic line bundle admits a Hermitian metric whose Chern connection curvature satisfies that is a positive real (1, 1)-form. In a local holomorphic local frame with squared length , the local formula for the Chern connection on a line bundle gives , so positivity means is positive definite. Its closedness makes a Kähler form; use this form to define the operators below.
Let , choose a Hermitian metric on , and equip with the tensor-product metric. The curvature of a tensor product connection givesOn -valued zero-forms, the Lefschetz commutator is . Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite such that at every point. For a holomorphic section of , and by degree, so integrating the identity givesChoose an integer with . Then . The threshold depends on the fixed bundle , as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 118 4 b Solution Created 2026-10-03 Updated 2026-10-05
Represent the middle-degree class by its unique harmonic differential form . The Lefschetz commutator on degree is , so at the operators commute. Since the adjoint Lefschetz operator is the adjoint of the Lefschetz operator of a Kähler manifold,Both and are harmonic. A harmonic form represents zero in de Rham cohomology exactly when the form itself is zero. Consequently if and only if , which by the norm identity is equivalent to , hence to . ThereforeThis identifies the two middle-degree descriptions of a primitive differential form on a Kähler manifold at the cohomology level.