Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 22 3 Solution Created 2026-10-03 Updated 2026-10-14
On a complex manifold, a divisor on a complex manifold is a locally finite formal sum , where ranges over irreducible complex analytic hypersurfaces and . The local algebra needed here is that the local ring , identified in coordinates with convergent power-series germs, is a regular local ring and a unique factorization domain. In particular, each height-one prime is generated by one irreducible germ, and a nonzero meromorphic germ is a quotient of holomorphic germs. Reduced local equations of the same hypersurface differ by a holomorphic unit. These facts include local branches when a globally irreducible hypersurface has several branches at a singular point.
The order of a meromorphic function along a hypersurface is the exponent of its reduced local equation in the quotient's factorization. Equivalently, at a generic smooth point of , write , with a transverse local equation of and a holomorphic unit; then . Changing a reduced equation multiplies it by a unit and leaves unchanged. These local orders agree on overlaps and are constant along the generic smooth locus of an irreducible hypersurface. A nonzero global meromorphic function therefore defines its principal divisor on a complex manifoldPositive coefficients record zeros and negative coefficients record poles.
A local defining function for on is a nonzero meromorphic function whose orders there equal the coefficients of . It can be constructed as the product of local reduced hypersurface equations to the prescribed integer powers. On an overlap, has no zero or pole along any hypersurface. The unique factorization domain property then makes it a nowhere-zero holomorphic function, after choosing representatives of the germs. The holomorphic line bundle associated to a divisor is obtained by gluing holomorphic frames usingThese ratios satisfy the cocycle condition, and the expressions agree on overlaps. They form the canonical meromorphic section of a holomorphic line bundle with divisor .
If globally, is a holomorphic unit on , and is a global nowhere-zero holomorphic section. Hence is holomorphically trivial. Conversely, if has a nowhere-zero global holomorphic frame , write with holomorphic units. The coefficients of the canonical meromorphic section of a holomorphic line bundle agree on overlaps and give a nonzero global meromorphic function . Since multiplication by a unit does not change orders, . Therefore
For a smooth hypersurface , the adjunction formula isIndeed, the holomorphic conormal sequence has first term ; taking its determinant yields , which gives the formula. Here the canonical bundle is the top exterior power of the holomorphic cotangent bundle.
Apply this first to . Its degree- equation is a section of , so . Since , we obtain . The restricted second equation cuts out in and has associated bundle . A second application of adjunction formula gives the canonical bundle of a regular projective complete intersection
It remains to prove that a nonzero exponent cannot yield a trivial holomorphic line bundle on . Choose a connected component and a hyperplane meeting but not containing it. Such a hyperplane exists: a positive-dimensional compact projective submanifold cannot lie entirely in an affine chart, since all its affine coordinate functions would be global holomorphic functions and hence constant. A hyperplane through one point and not another consequently meets properly. Its equation gives a holomorphic section of that has both a zero and a nonzero value. For , the th power has the same two properties. Were trivial, this section would be represented by a global holomorphic function, contradicting its constancy on compact connected . For , triviality would imply triviality of its positive dual power. This proves nontrivial hyperplane powers on a compact projective submanifold and henceThe degrees are positive because the two equations define a hypersurface and a surface of the specified dimensions.
The calculation uses “non-singular” in the usual smooth complete-intersection sense: the equations define reduced smooth divisors, with independent differentials along . If only their reduced zero sets are required to be smooth, polynomial multiplicities cannot determine the canonical bundle. For example, and have the same smooth quartic surface in the hyperplane for every . That surface has trivial canonical bundle, while the displayed polynomial degree sum is . Thus unreduced equations must not be substituted into the adjunction formula for the reduced surface.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 19 2 Solution Created 2026-10-03 Updated 2026-10-14
On a complex manifold, the exterior derivative splits by type as . The conjugate Dolbeault operator raises holomorphic degree by one and the Dolbeault operator raises antiholomorphic degree by one. Their identities areUse the convention for the d c operator; then . The alternative convention with a factor changes this last factor only. Complex conjugation of differential-form type interchanges and without reversing the wedge order. Thus, for a real form ,The resulting form is real, independently of the degree or type components of .
The Dolbeault cohomology isIn degree there is no incoming image, and the kernel consists of global holomorphic functions. On a compact connected , attains a maximum. In a coordinate polydisc around a maximizing point, the one-variable maximum modulus principle applied successively along the coordinate directions makes constant on a smaller polydisc. Alternatively the same argument shows the maximum set is open; it is closed and nonempty, so connectedness makes it all of , and the local constancy then makes globally constant. Consequently
A holomorphic p-form on a complex manifold is a form of type with holomorphic coordinate coefficients, equivalently one killed by . For degree , it is -closed: its vanishes by holomorphicity and its has impossible bidegree . Its conjugate is also closed.
The key positivity behind nonzero holomorphic top forms are not exact is explicit. If , complex orientation givesThis is nonnegative everywhere and positive on some open set when , so its integral on the compact manifold is strictly positive. If , closedness of and the Stokes theorem would giveHere Stokes is used on a compact oriented manifold without boundary, and complex-valued forms are handled by linearity. The contradiction proves a nonzero holomorphic top form is not -exact. For dimension zero the same nonexactness follows directly from the absence of degree forms.
Now take a compact complex surface, with no Kähler metric assumed. For a holomorphic one-form , the form has type and is holomorphic, since . It is exact. The top-form result forces , proving closedness. This establishes holomorphic forms on a compact complex surface are closed by an argument which does not require a Kähler identity.
If a holomorphic one-form were exact, write for a global smooth complex function. Comparing types gives . The preceding compactness argument makes constant on every connected component, hence . Holomorphic two-forms are closed and nonexact unless zero by the top-form result. Holomorphic zero-forms are constant on each component, and a nonzero zero-form is never exact because there are no negative-degree forms. Degrees above two contain only zero forms. Therefore every holomorphic form on a compact complex surface is closed, and a nonzero one is never exact.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 13 1 Solution Created 2026-10-03 Updated 2026-10-14
On the polydisc, put and defineEach integral is along the straight coordinate segment, which remains in the polydisc. Holomorphic dependence on parameters and differentiation under the integral show that is a holomorphic function. For a fixed , terms with do not depend on , and the term contributes . If , use the assumed symmetry to obtainThese differences telescope. ThereforeThis is the coordinate-integral primitive on a polydisc.
For a closed holomorphic one-form , closedness is exactly . The preceding construction plus gives a local holomorphic function with . The kernel of on the structure sheaf of a complex manifold consists of locally constant functions: holomorphicity already gives , and then implies . Thus the holomorphic Poincaré lemma gives the short exact sequence of sheavesThe first object is the complex constant sheaf, and surjectivity is local, on stalks. It does not say that every global closed holomorphic one-form has a global primitive.
Now assume compactness. The constancy assertion is for a connected complex manifold; otherwise it holds on each connected component. A smooth complex-valued with is a pluriharmonic function. Restricting to each complex coordinate line makes it a harmonic function of one complex variable. Choose where has its global maximum and a coordinate polydisc about . The allowed harmonic maximum principle makes the first coordinate-line restriction constant. Every point on that line still has the same maximal modulus and the same value, so the second coordinate may then vary without changing . Continue through all coordinates to conclude that equals on the whole polydisc. The set where it has this value is closed and, by the same argument at each of its points, open. Connectedness therefore gives constant. In particular global holomorphic functions on a compact connected complex manifold are constant.
For the top-degree pairing, write locally . With the complex orientation,It is everywhere nonnegative and is positive on an open set if is not identically zero. HenceThe phase factor is necessary: the unnormalized integral need not be a positive real number. This is the integral pairing of complex top forms.
If is a holomorphic differential form of degree , then has type and . The graded Leibniz rule and Stokes theorem giveThe positive pairing just proved forces . Thus every holomorphic -form on a compact complex -manifold is closed, without a Kähler metric assumption.
On a compact complex surface, this proves closedness of all holomorphic one-forms. Sending such a form to its de Rham cohomology class defines the required inclusion. Indeed, if , its type gives , so is a global holomorphic function and is constant. Therefore .
To prove the zero intersection, let be holomorphic one-forms and suppose . There is a smooth complex-valued with . Its type components are and . Closedness of gives , hence . Compact pluriharmonic rigidity now makes constant and . The two spaces are therefore linearly independent in de Rham cohomology, and conjugation preserves their dimensions:For the other bound, take the long exact sequence in sheaf cohomology of the short exact sequence above. The map is an isomorphism, componentwise. Since every global holomorphic one-form is closed, the relevant portion isIdentify the middle term with first de Rham cohomology, and the last with by the Dolbeault theorem. Taking dimensions givesThese are the Betti and Hodge bounds for compact complex surfaces.
A strict example is the scalar Hopf surface . Polar coordinates identify it smoothly with : the dilation fixes the angular coordinate and shifts by . Its first Betti number is one, since its fundamental group is . The proved lower bound then forces , and the upper bound forces . Thus on this compact complex surface; no unsupported cohomology computation is needed.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 17 1 Solution Created 2026-10-03 Updated 2026-10-07
The divisor determines the bundle through its local equations. Choose an open cover on which has a reduced holomorphic defining equation , using on sets disjoint from . Because is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With , its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is . The section is holomorphic and vanishes to order one along . Replacing by , with a holomorphic unit, replaces the frame by without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.
For a complex manifold of dimension , the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let be the complex tautological line bundle, so that is the hyperplane line bundle. At a line , the tangent space is . Tensoring the tautological quotient sequence by gives the Euler sequence on complex projective spaceTaking the top exterior power gives ; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.
The proposed sections over form a sheaf: compatible maps into glue uniquely, and their projections remain . Addition and multiplication by holomorphic functions are defined in the fibres of , making this a sheaf of -modules. If is a holomorphic local frame of on , every section over has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of composed with , so the corresponding pullback vector bundle is precisely . This proves invertibility locally and compatibility globally.
For the local blowup of a complex manifold at a point, work on the projective chart and writeAll the incidence equations reduce to . Therefore gives a holomorphic coordinate chart isomorphic to . On an overlap with ,which is a biholomorphism where . The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension . In each chart the exceptional fibre is , so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is , which is holomorphic; the projection is therefore a biholomorphism outside .
To find the canonical bundle, pull back the nowhere-zero top form on . In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of with divisor on a complex manifold exactly : it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor identifies its holomorphic line bundle with , by sending the local generator of to the nowhere-zero frame obtained by dividing the section by . ConsequentlyThis also includes , when the local map is an isomorphism and the exponent is zero.
For a general complex manifold, choose a coordinate neighbourhood of , with sent to , and replace by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to along using the preceding biholomorphism. The resulting charts give a complex manifold and a holomorphic map ; the exceptional fibre is and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over inside the local blowup. The atlas is second countable as well.
The top exterior power of the differential defines a global holomorphic bundle map . Equivalently, it is a section of . The same local Jacobian determinant has vanishing order on , and the differential is invertible elsewhere. Applying the holomorphic line bundle associated to a divisor construction gives the canonical bundle formula for a point blowup
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 17 2 a Solution Created 2026-10-03 Updated 2026-10-06
A rank- holomorphic vector bundle is a complex manifold with a holomorphic projection to and local holomorphic trivializations that are complex-linear on each fibre. Its transition maps have the form with holomorphic satisfying the cocycle identities.
The holomorphic Picard group consists of isomorphism classes of holomorphic line bundles, with tensor product as multiplication, the trivial line bundle as identity and the dual bundle as inverse. The classification on the complex projective line givesThe inverse is degree; is the positive generator.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 2 c Solution Created 2026-10-03 Updated 2026-10-14
We prove the Bott-Chern Poincaré lemma on a polydisc . Let and , with of type . The allowed Dolbeault-Poincaré lemma first gives , where has type . Since , repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form has type , is holomorphic because its Dolbeault operator vanishes, and is -closed. The holomorphic Poincaré lemma on a polydisc gives a holomorphic form of type with . This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincaré lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take .
Now is -closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives , since its holomorphic degree is positive. Moving backwards, if the last modified form is , thenApply the conjugate Dolbeault-Poincaré lemma again to obtain . The modification does not change its Dolbeault operator, so this process continues down to . In the case , subtract at this last step instead. Finally,Thus .
This vanishing fails on general complex manifolds. On the complex projective line, let be its Fubini-Study form. It is a -closed -form with . If it were , it would be the exact differential form , whose integral is zero by Stokes theorem. Therefore .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 3 a Solution Created 2026-10-03 Updated 2026-10-06
Order the index set of the open cover . For the sheaf of abelian groups , the Čech cochain group isIt is a product, not a finite-support sum. The Čech coboundary isEvery term in occurs twice with opposite signs, so . The Čech cohomology of the cover is . Refinements induce maps on these groups; different choices of refinement indices give maps related by a cochain homotopy, hence the same map on cohomology. Taking the direct limit over covers givesOne may use locally finite covers, since a complex manifold is a paracompact space and these covers are cofinal. In degree zero this recovers the group of global sections by the sheaf gluing axiom.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 5 d Solution Created 2026-10-03 Updated 2026-10-06
Choose a positive Hermitian metric on . Its curvature form is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold with for every . We use this to prescribe a finite principal part of a meromorphic section.
Fix a smooth cutoff supported inside the given coordinate chart and equal to one near . Let . On the punctured manifold define inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is -closed, either by the square-zero identity or because there are no -forms in complex dimension one. The vanishing of , together with the Dolbeault theorem, therefore gives a global smooth section with .
The section is holomorphic on . Near , , so with holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of . The threshold is determined by the fixed bundle and , and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order , since the Dolbeault cohomology obstruction is the same for every such cutoff construction.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 118 1 a Solution Created 2026-10-03 Updated 2026-10-06
The complex structure splits the complexified cotangent bundle into its and parts. Thus the space of smooth differential forms of type (p, q) isHere denotes smooth sections of a vector bundle; exterior multiplication identifies the displayed bundle with a subbundle of the complexified exterior algebra. In holomorphic coordinates, such a differential form of type (p, q) has the expressionwhere the coefficients are smooth and both multi-indices are increasing. Negative bidegrees, or bidegrees exceeding the complex dimension, give the zero space.
On a complex manifold, the exterior derivative has just two type components, . The conjugate Dolbeault operator and the Dolbeault operator are, respectively,Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. The two operators are the type projections of the exterior derivative, raising and , respectively.
For a holomorphic map , its differential is a complex-linear map, so the pullback of a differential form sends covectors to covectors and covectors to covectors. In coordinates on ,The pullback of a differential form commutes with the wedge product of differential forms, so it preserves the counts of these factors. ConsequentlyIt also commutes separately with the Dolbeault operator and the conjugate Dolbeault operator, because it commutes with and preserves type.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 118 5 d Solution Created 2026-10-03 Updated 2026-10-05
Let be the sheaf of closed holomorphic one-forms, namely . The local exactness proved in part (b) gives the short exact sequence of sheavesIts long exact sequence in sheaf cohomology beginsThe first arrow is an isomorphism: on a compact complex manifold, a global holomorphic function is constant on each connected component, just as a global section of the constant sheaf is. Thus is injective. By part (c), every global holomorphic differential form of degree one is closed, so .
Concretely, for a holomorphic differential form of degree one choose local holomorphic primitives . The connecting class is represented by the locally constant Čech cocycle on overlaps. The following arrow is induced by the inclusion of the constant sheaf into the sheaf of holomorphic functions. Exactness is inherited from the long exact sequence, giving
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 118 1 a Solution 2026-10-03
A complex manifold of complex dimension is a Hausdorff, second-countable topological manifold with charts to open subsets of whose transition maps are biholomorphic. A rank- holomorphic vector bundle is a complex vector bundle with local trivializations whose transition matrices are holomorphic maps to .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 118 3 c Solution 2026-10-03
A holomorphic line bundle is ample when some positive tensor power is very ample, so its sections define an embedding into Complex projective space. It is a positive holomorphic line bundle when it has a Hermitian metric with positive Chern curvature . The Kodaira embedding theorem says that a compact complex manifold carrying a positive holomorphic line bundle is projective; sufficiently high tensor powers give a holomorphic embedding into projective space.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 24F a Solution Created 2026-09-24 Updated 2026-09-29
LetIts complex partial derivatives areThere is no point of at which both vanish. Indeed, if , then and giveNeither nor can be zero because . Substitution into gives , whereas substitution into gives ; these equations are incompatible.
For every , at least one partial derivative is nonzero. If , the holomorphic implicit function theorem expresses the curve near as , and the restriction of the -projection is a holomorphic coordinate chart. If , use the -projection instead. On overlaps, the transition functions are restrictions of holomorphic coordinate functions and are holomorphic. These charts define an atlas making a one-dimensional complex manifold, hence a Riemann surface.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 118 1 Solution Created 2026-09-28 Updated 2026-10-03
An almost complex structure on a smooth manifold is a smooth bundle endomorphism satisfying . On the complexified tangent bundle, has the eigenbundle decompositionwith eigenvalues and . A differential form of type (p, q) is a section of . The operators and are the type- and type- components of the exterior derivative.
We prove the three stated conditions are equivalent. If have type and has type , the formula for the exterior derivative givesThus closure of under the Lie bracket is equivalent to the vanishing of the component of for every -form. Complex conjugation of differential-form type gives the corresponding vanishing of the component on -forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)(ii).
Under (ii), the component of is , proving (iii). Conversely, for of type ,If this vanishes for every smooth complex-valued , then , so (iii) implies (i). These conditions define an integrable almost complex structure.
For a complex manifold, a holomorphic chart defines and . The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by ; the definitions therefore glue and are independent of coordinates. Locally is spanned by the commuting fields , so it is closed under brackets and is integrable. This is the almost complex structure induced by a complex atlas.
We next prove the local Dolbeault-Poincaré lemma. Write a -closed -form on a slightly larger polydisc aswhere neither nor contains . Apply the supplied one-variable Cauchy-Green operator coefficientwise to , obtaining with . Then contains no , and says that its coefficients are holomorphic in and -closed in the first variables. Induction on , with the one-variable formula as the base case, makes this remainder -exact. Hence every -closed -form with is -exact on each bounded polydisc.
Finally, translation by leaves unchanged, so this form descends to . It is in fact exact on this noncompact complex cylinder, because the invariant function descends and satisfies
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 118 1 Solution 2026-09-28
Let be a complex manifold chart, with . The almost complex structure induced by a complex atlas isThus . On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by . The two coordinate definitions of therefore agree, so is globally well-defined.
The complexified cotangent bundle splits into the - and -eigenspaces of , locally spanned by and . A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends to and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity
Let act on a -form byIt acts on a -form by . Therefore multiplies the component by and the component by , proving the d c operator formula
A holomorphic vector field is a holomorphic section of . On the affine chart , put . The projection isDirect differentiation givesIf is linear and homogeneous, the pushed-forward coefficients are consequently when , and when . They are holomorphic on .
More intrinsically, is a linear vector field on . Its flow preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on is the field just computed. This proves extension across the other affine charts.
Finally choose distinct complex numbers and the diagonal fieldIts projectivization vanishes at precisely when is proportional to , so is an eigenline. The distinct eigenvalues leave exactly the coordinate points. Hence has a holomorphic vector field with finitely many zeroes.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 118 2 Solution Created 2026-09-28 Updated 2026-10-14
An irreducible complex analytic hypersurface in a complex manifold is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface at is a holomorphic function on a neighbourhood such thatThe necessary local algebra is that the stalk is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives , and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.
A divisor on a complex manifold is a locally finite formal sum of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small , local defining functions give a meromorphic equation for . The quotients are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor , and gives its canonical meromorphic section with divisor .
The Euler sequence on complex projective spaceimplies . Taking the dual determinant yields the canonical bundle of complex projective spacewhere is a hyperplane divisor.
The hypotheses on the homogeneous polynomial say thatis a smooth projective hypersurface of degree , so its divisor line bundle is . The adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.
Now fix an isomorphism and regard and as holomorphic sections of the same holomorphic line bundle. They have no common zero because . Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor . Thus and , both with multiplicity one. The degree of a holomorphic map is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.