Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 18 / 3

Order the index set of the open cover $\mathcal{U}=(U_i)$. For the sheaf of abelian groups $\mathcal{F}$, the Čech cochain group isIt is a product, not a finite-support sum. The Čech coboundary isEvery term in ${\delta }^{2}c$ occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^p(\mathcal{U},\mathcal{F})=\ker \delta /\mathrm{im}\delta$. 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.

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 $E$ 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 $j_{\ast }\mathcal{O}(E{|}_{X_0})$ with $\mathcal{O}(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher sheaf cohomology.

Here is an explicit higher-degree obstruction. Take $X={\mathbb{P}}^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne 0\}$ and $V=\{y\ne 0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are ${\mathbb{C}}^2$, and the overlap is ${\mathbb{C}}^{\ast }\times \mathbb{C}$. For every integer $m\ge 2$, the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}{b}_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\ge 2$, are linearly independent.

These classes remain nonzero in Čech cohomology of $X_0$, 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 $H^1(X_0,\mathcal{O})$. Concretely, a partition of unity gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common Dolbeault operator defines a global $(0,1)$-form. If that form were $\bar{\partial }$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus $H^1(X_0,\mathcal{O})$ is infinite dimensional. In contrast, $H^1({\mathbb{P}}^2,\mathcal{O})$ 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 $X={\mathbb{P}}^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected ${\mathbb{P}}^1$ are constant by the maximum modulus principle, whereas $X_0\cong \mathbb{C}$ has nonconstant entire functions such as $z$. Therefore the restriction map is not onto even in degree zero.

The complex tautological line bundle on ${\mathbb{P}}^1$ has fibre at a line $\ell \subseteq {\mathbb{C}}^2$ equal to $\ell$ itself:Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$, $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ for $k>0$, $\mathcal{O}(0)$ to be trivial, and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ for $k<0$. These are the tensor powers of the hyperplane line bundle.

On the two given charts, use the holomorphic local frames $e_0(w)=(1,w)$ and $e_1(z)=(z,1)$ of $\mathcal{O}(-1)$. The maps $([1:w],a)\mapsto ([1:w],a(1,w))$ and $([z:1],b)\mapsto ([z:1],b(z,1))$ are its trivializations. On the overlap $z=1/w$,Thus the frame-transition factor from $e_0$ to $e_1$ is $w^{-1}$, while the fibre-coordinate transition from chart 0 to chart 1 is $w$. Specifying both avoids an inverse-convention ambiguity.

For $\mathcal{O}(k)$, let $e_i^{(k)}$ be the induced frames; they satisfy $e_1^{(k)}=w^k{e}_0^{(k)}$. Identifying overlap sections using $e_0^{(k)}$, the Čech cochain groups and Čech coboundary for $k\ge 0$ areHere $\mathcal{O}(\mathbb{C})$ means entire functions in the coordinate of the corresponding chart. Global sections are $\ker \delta$, so $a(w)=w^{k}b(1/w)$, or $b(z)=z^{k}a(1/z)$. Expanding the entire function $a(w)={\sum }_{j\ge 0}{a}_j{w}^j$ shows that $b$ is holomorphic at zero exactly when $a_j=0$ for $j>k$. Thus a global section is determined by a polynomial of degree at most $k$, with basis $1,w,\dots ,w^k$ in the chart-0 frame. Therefore $\boxed{\dim H^0({\mathbb{P}}^1,\mathcal{O}(k))=k+1(k\ge 0)}$.