Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 18 / 2

We use complex-valued smooth differential forms. The sheaf of smooth differential forms is ${\mathcal{A}}^k(U)=\Gamma (U,{\bigwedge }^k(T^{\ast }X\otimes \mathbb{C}))$, with the usual restriction maps. Dualizing the type decomposition of the complexified tangent bundle and taking exterior powers decomposes this bundle into the summandsTheir smooth sections form the sheaf of differential forms of type (p, q) ${\mathcal{A}}^{p,q}$. Locally a section is a sum of $f_{I,J}d{z}_I\wedge d{\bar{z}}_J$ with $|I|=p$, $|J|=q$ and smooth coefficients. Holomorphic transition maps preserve types, so the local decompositions agree globally. HenceThe exterior derivative splits as $d=\partial +\bar{\partial }$, with bidegrees $(1,0)$ and $(0,1)$. Its square being zero gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault cohomology is thereforeForms in negative or out-of-range bidegrees are understood to be zero.

For a differential form of type (p, q), $d\alpha =0$ is equivalent to both $\partial \alpha =0$ and $\bar{\partial }\alpha =0$, because the two resulting types are distinct. Moreover $\partial \bar{\partial }\beta$ has type $(p,q)$ when $\beta$ has type $(p-1,q-1)$, andby the square-zero and anticommutation identities. Thus the denominator defining Bott-Chern cohomology is a vector subspace of its numerator, making the quotient well defined.

Complex conjugation of differential-form type sends a $d$-closed $(p,q)$-form to a $d$-closed $(q,p)$-form. For a representative in the denominator,The minus sign does not change the denominator vector subspace. Conjugation consequently induces a conjugate-linear bijection $[\alpha ]\mapsto [\bar{\alpha }]$ between the two Bott-Chern cohomology spaces; applying it twice is the identity. Equivalently, $\boxed{H_{BC}^{p,q}(X)\cong \overline{H_{BC}^{q,p}(X)}}$ as complex vector spaces.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincare lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincare lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

A $d$-closed differential form of type (p, q) is $\bar{\partial }$-closed, and a Bott-Chern cohomology coboundary is a Dolbeault cohomology coboundary becauseHence taking the same representative defines a complex-linear map $\varphi :H_{BC}^{p,q}(X)\to H_{\bar{\partial }}^{p,q}(X)$.

If $X$ is compact and Kähler, the Dolbeault Hodge decomposition supplies a $\bar{\partial }$-harmonic representative $\alpha$ of every Dolbeault cohomology class. The scalar Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}$, so $\alpha$ is also $d$-harmonic. Consequentlyand $d\alpha =0$. It therefore defines a Bott-Chern cohomology class mapping to the original class. Thus $\boxed{\varphi \text{ is surjective on a compact Ka¨hler manifold}.}$

For clarity, the scalar Kähler Laplacian identity follows from the Kähler identities: they make the mixed anticommutators $\{\partial ,{\bar{\partial }}^{\ast }\}$ and $\{\bar{\partial },{\partial }^{\ast }\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives ${\Delta }^{\prime }={\Delta }^{\prime \prime }$. Expanding ${\Delta }_d$ yields ${\Delta }^{\prime }+{\Delta }^{\prime \prime }=2{\Delta }^{\prime \prime }$.