Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 3 b Solution Created 2026-10-03 Updated 2026-10-06
The printed assertion for every cohomological degree is false. The valid conclusion supplied by the stated Hartogs extension theorem is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the holomorphic vector bundle is trivial. A holomorphic section on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by Hartogs extension theorem when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, givingThis also identifies the direct image sheaf with for the inclusion . It does not identify higher sheaf cohomology.
Here is an explicit higher-degree obstruction. Take , remove , and take the trivial line bundle. Cover by and . On use coordinates , ; on use , . Both charts are , and the overlap is . For every integer , the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary with entire on their charts. Taking the coefficient of , the term from has only nonnegative powers of , whereas the term from has the form and only powers at most . Neither can provide the coefficient . Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes , , are linearly independent.
These classes remain nonzero in Čech cohomology of , rather than merely of this cover. The low-degree Mayer-Vietoris sequence for sheaf cohomology injects the quotient of overlap sections by the two chart-section groups into . Concretely, a partition of unity gives smooth with ; their common Dolbeault operator defines a global -form. If that form were -exact, subtracting its global smooth primitive would make holomorphic and split . The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus is infinite dimensional. In contrast, is finite dimensional by compact Dolbeault Hodge decomposition for the Fubini-Study form. Hence the two degree-one groups cannot be isomorphic, disproving the printed all-degree request already in complex dimension two.
In complex dimension one, even the degree-zero conclusion fails. Take , remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected are constant by the maximum modulus principle, whereas has nonconstant entire functions such as . Therefore the restriction map is not onto even in degree zero.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 5 a Solution Created 2026-10-03 Updated 2026-10-06
For a holomorphic vector bundle , the sheaf of vector-bundle-valued differential forms iswhere sections are smooth and restrictions are the usual ones. In a holomorphic local frame of , the Dolbeault operator acts coefficientwise and satisfies . Its Dolbeault cohomology with values in a holomorphic vector bundle is the cohomology of the global section complex .
The bundle-valued Dolbeault theorem identifies this with sheaf cohomology:Here is the sheaf of holomorphic differential forms, and is the sheaf of holomorphic sections of a vector bundle. Equivalently, the complexis a fine sheaf resolution, by the local Dolbeault-Poincaré lemma and smooth partitions of unity. In particular computes ; for the holomorphic differential-form factor must be retained.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 18 5 b Solution Created 2026-10-03 Updated 2026-10-06
The Kähler form and the Hermitian metric give the inner product on vector-bundle-valued differential forms, using volume . Define the formal adjoint and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from . On compact , the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for , whose square is zero; it does not require the full Chern connection to be flat.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 113 4 ii Solution Created 2026-10-03 Updated 2026-10-06
For the closed inclusion , the direct image sheaf has stalk at and zero stalk at points outside : outside the closed set there is an open neighbourhood disjoint from . Hence is exact, since exact sequences of sheaves are detected on stalks.
Take a flasque resolution . Direct image preserves flasqueness, because restrictions are the restrictions on inverse-image opens. Exactness makes a flasque resolution of . MoreoverThe complexes computing sheaf cohomology are identical. ThereforeThis is sheaf cohomology under a closed inclusion; the closedness hypothesis is used in the exactness of direct image.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 118 4 b Solution Created 2026-10-03 Updated 2026-10-06
The two opens are and ; their intersection is . There is just one degree-one term in the Čech cochain complex, and no degree-two term. The Čech coboundary sends to , soThe denominator denotes the sum of the restricted function spaces.
To make the quotient explicit, expand a holomorphic function on the intersection in a normally convergent two-variable Laurent series,The terms with extend to . Of the remaining terms, those with extend to . Both subseries converge normally on their stated domains, by the coefficient bounds from the Cauchy integral formula. The unique remaining representative is the doubly negative partNo nonzero series of this form lies in the denominator, since a function on has no negative exponents and a function on has no negative exponents.
Writing , these representatives are exactly with an entire holomorphic function on . To see that is entire, integrate for the coefficients on arbitrarily small product circles: . Choosing gives absolute convergence for , for every finite pair . Conversely every such entire supplies a normally convergent representative on the intersection. Therefore the quotient consists of convergent doubly negative Laurent series, not merely finite Laurent polynomials:This gives the holomorphic first cohomology of punctured complex two-space; for instance represents a nonzero class.
For the comparison with sheaf cohomology, and their intersection are Stein manifolds. They can be realized as closed complex submanifolds of affine complex spaces by adding equations for their nonzero coordinates. Cartan theorem B makes their higher cohomology with coefficients in the sheaf of holomorphic functions vanish. Thus the cover is acyclic for , and the acyclic cover theorem givesThe individual cover members are Stein; their union has the nonzero cohomology just computed.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 113 4 ii Solution Created 2026-10-03 Updated 2026-10-05
Write , where , and , with on the overlap. Since is a quasi-coherent sheaf, it is associated to a -module on . The preimages under the open inclusion of are , all affine schemes. Thus is an affine morphism and its direct image sheaf is quasi-coherent by direct image of a quasi-coherent sheaf under an affine morphism.
The two-chart cover is an acyclic cover by vanishing of quasi-coherent cohomology on an affine scheme, so the acyclic cover theorem identifies its Čech cohomology with sheaf cohomology. Its only potentially nontrivial positive-degree differential isThis is surjective because of the second summand. There are no normalized Čech terms in degree two or higher for a two-open cover, soThis is the acyclic direct image from an affine chart of the projective line. The pushforward need not itself be a flasque sheaf; the acyclic affine cover is what justifies this computation.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 113 3 a Solution Created 2026-10-03 Updated 2026-10-05
First consider the preliminary cohomological facts. The exam's term flabby means flasque sheaf: every restriction map is surjective. For open sets , lift a section of to using the given surjectivity when is flasque. Extend that lift to using the flasque sheaf , then project to . This proves that is a flasque sheaf.
One construction of a flasque resolution starts with the sheaf , with pointwise restriction. It is a sheaf of abelian groups with surjective restrictions, and sending a section to all its germs embeds in it. Repeating on the cokernels constructs an exact sequencewith each flasque. Define sheaf cohomology by the cohomology of the cochain complex of global sections of such a resolution; the groups are independent of the chosen flasque resolution. If is itself flasque, the quotient result just proved makes every successive cokernel flasque. Applying the assumed surjectivity of sections to each successive short exact sequence of sheaves proves exactness of the global-section complex in positive degrees. Thus for .
Now let be the function field of the smooth algebraic curve. Its divisor class group isHere is the discrete valuation of the discrete valuation ring , and consists of the principal divisor elements . These sums have finite support: on a finite affine cover, represent as a fraction of regular functions; each nonzero numerator or denominator has only finitely many zeros on a curve, since a proper closed subset of a Noetherian one-dimensional irreducible space is finite.
For a divisor on an algebraic curve , define the line bundle associated to a divisor byIn this expression the valuation condition applies to nonzero . If is a uniformizer, the stalk is . Shrinking around removes all other zeros and poles of and all other points in the support of , so this also gives an actual local generator. Thus is a locally free sheaf of rank one. Multiplication gives an isomorphismon every stalk. If , then , so the construction factors through a homomorphism .
It is injective: if is trivial, an isomorphism from supplies a nonzero global rational generator . At each closed point, generates , giving , hence .
It is surjective: take a line bundle and a nonzero rational section , obtained by choosing a frame on any nonempty trivializing open set. If is a local frame, write with . Since changes of frame are units, is independent of the frame near . A finite trivializing cover shows that these numbers have finite support. Set . The map gives , because at the condition is exactly regularity of . Replacing by changes by , so this construction is well defined on classes. This proves the divisor class group and Picard group of a smooth curve identification
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 113 4 Solution Created 2026-10-03 Updated 2026-10-05
The resolution principle for sheaf cohomology states that an exact sequencewith for every computes as the cohomology of . In particular one may use a flasque resolution.
An affine morphism is one for which the inverse image of every affine open subset of is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution . Each direct image sheaf is flasque, because restriction on is restriction on . On an affine open , the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is . Checking exactness on this affine basis shows that is a flasque resolution of . The global-section complexes are identical, proving cohomology under an affine morphism:
For projective space, put with its usual grading and . Construct the twisting sheaf on projective space by setting its sections on to be the degree- part , viewed as a module over . It is free of rank one with generator , even when is negative. On overlaps the generators differ by the unit , so these modules glue to a line bundle .
Assume first . These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with . Since is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every has no nonconstant denominator. Thus the intersection is . Counting monomials by stars and bars givesHere . The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For , every twist on the single point is trivial, so for every and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for .
The standard affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree . In particular for .
Let , with inclusion . Multiplication by gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to . Since a closed immersion is an affine morphism, the first part identifies the cohomology of with that on .
For , the long exact sequence in sheaf cohomology and giveStarting from the given vanishing for sufficiently large and descending gives for and for .
Now let , and assume the formulas in dimension . Restrictionis surjective for every : for it is polynomial restriction, and for both groups are zero since . The long exact sequence in sheaf cohomology therefore makes injective. For , dimension induction gives , so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves for every and .
The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on . Hence . Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is , obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 118 2 a Solution Created 2026-10-03 Updated 2026-10-05
Choose a flasque resolution of the sheaf of abelian groups ,A flasque sheaf has surjective restriction maps; such resolutions exist for every sheaf of abelian groups. Apply the global section functor to obtain a cochain complex. The sheaf cohomology groups areFor , the denominator is zero and . Different resolutions give canonically isomorphic groups; the resolution principle for sheaf cohomology is what permits other acyclic resolutions to compute these same groups.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 118 2 b Solution Created 2026-10-03 Updated 2026-10-05
Let be the additive sheaf of complex-valued smooth functions, and its multiplicative sheaf of nowhere-zero functions. The smooth exponential sequence isIts kernel is the integer-valued locally constant functions, namely the integer constant sheaf. It is surjective on stalks because a nowhere-zero smooth function has a smooth logarithm on a sufficiently small neighbourhood.
The additive smooth-function sheaf is a fine sheaf, using a partition of unity, and the smooth manifold is a paracompact space. Its positive-degree sheaf cohomology vanishes. The long exact sequence in sheaf cohomology consequently gives an isomorphism
To connect this with bundles, trivialize a smooth complex line bundle over an open cover. Its transition functions form a multiplicative Čech cocycle, satisfying . Changing trivializations changes it by a Čech coboundary. Conversely, any such cocycle glues trivial line bundles, and cohomologous cocycles give isomorphic bundles. Thus the isomorphism classes are .
The connecting isomorphism is the First Chern class. Locally choose logarithms ; on triple intersectionsrepresents that class. Combining the gluing classification with gives the requested bijection:
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 1 iv Solution Created 2026-10-03 Updated 2026-10-05
By asymptotic Riemann–Roch and the preceding bounds for higher sheaf cohomology,A nef divisor satisfies : for a fixed ample divisor , the divisors are ample for positive rational , and continuity of the intersection product gives the inequality as . If eventually, the left side is nonpositive. Dividing by forces , since and . The same identity then gives .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 1 v Solution Created 2026-10-03 Updated 2026-10-05
First suppose is integral. Induct on its dimension of a scheme. The previous two parts handle all and also handle when eventually vanishes. Otherwise choose with a nonzero global section of . On an integral variety it defines an effective Cartier divisor , possibly empty, and multiplication by the section givesThus . For , the induction hypothesis bounds the last term by ; summing on each residue class modulo gives . For , is zero-dimensional and its positive-degree sheaf cohomology vanishes, so the same recurrence is bounded.
For a general projective scheme, a nonzero section can be a zero divisor, so that argument requires an additional step. The general cohomology growth for nef twists supplies it: for every coherent sheaf with support dimension ,Its proof uses Fujita vanishing, an ample section avoiding the associated points of , and induction on support dimension. Taking gives the required estimate for every , including nonreduced and reducible schemes; degrees above vanish.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 2 iv Solution Created 2026-10-03 Updated 2026-10-05
A connected integral projective variety has only constant regular functions, so . By Serre duality and , . The assumed first sheaf cohomology vanishes. Therefore the Euler characteristic of a coherent sheaf isThese hypotheses characterize the K3 surface setting used below.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 113 4 c Solution 2026-09-28
Write , , and . Since is an integral domain, no nonzero global section is supported only at the origin, soThe punctured affine plane has the affine cover . Its Čech cochain complex for the structure sheaf isConsequently
Because is affine, the higher sheaf cohomology of vanishes. The long exact sequence for local cohomology therefore givesand for . The nonzero group has the -basisThis computes the local cohomology of the affine plane supported at the origin in every degree.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 113 3 c Solution 2026-09-28
Let and cover the punctured affine three-space by the three principal opens . Every finite intersection is affine, so the acyclic cover theorem identifies sheaf cohomology with the cohomology ofThe augmented complex has zeroth cohomology and first cohomology zero. Its second cohomology iswith -basis represented by for . There are no higher Čech terms. Hence
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 113 3 b Solution Created 2026-09-24 Updated 2026-09-25
The complement is covered by the principal affine opens . Every finite intersection is again a principal affine open, so this is an acyclic cover for . Its Čech cochain complex has no degree- term because there is no intersection of distinct cover members. Therefore
Translate to the origin. The complement has the affine cover , and its second Čech cohomology isThe class of is nonzero, so . On the other hand, , and the first part with gives . Since sheaf cohomology is invariant under scheme isomorphism, the two complements are not isomorphic.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 113 3 b Solution Created 2026-09-24 Updated 2026-09-25
Write and . Since , the principal open subscheme contains , andCover by the two affine opens and , whose intersection is . The degree-zero part of the resulting Čech cohomology complex givesinside the fraction field of . The last equality follows because is a unique factorization domain and a rational function regular after localizing at both and has no possible prime factor left in its denominator.
The same affine cover is acyclic, so its degree-one Čech group computes sheaf cohomology and givesBefore localizing at , the quotienthas the -basisWriting with and , multiplication by is locally nilpotent on : for each negative monomial, a sufficiently high power of moves every term into . Hence acts invertibly on by a finite geometric series on each element. Localizing at therefore leaves unchanged, andThe displayed infinite basis proves that this vector space is infinite-dimensional.
Picard charts from nonspecial divisors 2026-10-14
A degree- line bundle with vanishing first sheaf cohomology has exactly one section up to scale, by Riemann-Roch theorem. Its zero divisor is unique and varies in families by cohomology and base change for line bundles on a curve. For a fixed divisor of degree , the open subset of where has vanishing first cohomology gives a chart in degree , by . Different charts are glued by the unique residual divisor of wherever that line bundle is again nonspecial. Normalizing the universal bundles along a fixed rational point gives their canonical compatible gluing and a representing scheme for the Picard functor of a curve.
Projective complete intersection Created 2026-10-05 Updated 2026-10-14
A projective complete intersection is a closed subscheme of a projective space defined by a homogeneous regular sequence of positive degrees. A length- sequence in gives dimension when . Sheafifying its successive quotient sequences gives exact restriction sequences of twisting sheaves on projective space, allowing computation of its sheaf cohomology. It need not be smooth or reduced.
Resolution principle for sheaf cohomology 2026-10-05
If is an exact sequence of sheaves of abelian groups on and for every and , then sheaf cohomology is computed by the cochain complex of global sections:In particular a flasque resolution computes sheaf cohomology. Acyclicity is required on the space on which cohomology is being computed.