Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 118 / 4

Let $\omega$ be the Kähler form associated with the metric. The Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator areHere the adjoint is taken with respect to the pointwise Hermitian inner product on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the Hodge star operator, $\Lambda ={\ast }^{-1}L\ast$.

Since $d\omega =0$, $dL=Ld$, so $L$ takes closed differential forms to closed differential forms and exact differential forms to exact differential forms. This already gives its action on de Rham cohomology. For both operators together, use the Kähler identities: $L$ commutes with the Hodge Laplacian, and taking adjoints shows that $\Lambda$ does too. Thus both preserve harmonic differential forms.

When $M$ is compact, the Hodge decomposition theorem identifies $H^k(M,\mathbb{C})$ with its space of harmonic $k$-forms. If $h_{\alpha }$ is the unique harmonic differential form representing $[\alpha ]$, defineThe resulting forms are harmonic and therefore closed, and uniqueness of $h_{\alpha }$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses harmonic differential forms as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.

Represent the middle-degree class by its unique harmonic differential form $h$. The Lefschetz commutator on degree $k$ is $[\Lambda ,L]=(n-k)\mathrm{id}$, so at $k=n$ the operators commute. Since the adjoint Lefschetz operator is the adjoint of the Lefschetz operator of a Kähler manifold,Both $Lh$ and $\Lambda h$ are harmonic. A harmonic form represents zero in de Rham cohomology exactly when the form itself is zero. Consequently $L[\alpha ]=0$ if and only if $Lh=0$, which by the norm identity is equivalent to $\Lambda h=0$, hence to $\Lambda [\alpha ]=0$. ThereforeThis identifies the two middle-degree descriptions of a primitive differential form on a Kähler manifold at the cohomology level.

The assertion is pointwise, so choose an oriented orthonormal real coframe $x_1,y_1,x_2,y_2$ compatible with the complex structure, with Kähler form $\omega =x_1\wedge y_1+x_2\wedge y_2$. Real $(1,1)$ forms have a four-dimensional space; imposing $\omega \wedge \alpha =0$ leaves the three-dimensional space spanned byThese are precisely the real primitive differential forms on a Kähler manifold of type $(1,1)$ in dimension two. Complexifying spans all complex primitive $(1,1)$ forms.

Using the complex orientation $x_1\wedge y_1\wedge x_2\wedge y_2$, the Hodge star operator exchanges $x_1\wedge y_1$ with $x_2\wedge y_2$, sends $x_1\wedge x_2$ to $-y_1\wedge y_2$, and exchanges $x_1\wedge y_2$ with $y_1\wedge x_2$. Thus $\ast A=-A$, $\ast B=-B$, and $\ast C=-C$. By complex linearity,This is the fact that primitive (1,1)-forms on a Kähler surface are anti-self-dual.

Use the usual connected-manifold convention for this assertion. Suppose, for contradiction, that $H^2(M,{\mathcal{O}}_M)=0$. The Dolbeault theorem gives $H^{0,2}=0$, and the Hodge decomposition theorem for compact Kähler manifolds and its conjugation symmetry give $H^{2,0}=0$. Hence every real class in $H^2(M,\mathbb{R})$ has a real harmonic differential form of type $(1,1)$ as its representative.

The intersection form is $Q([u],[v])={\int }_{M}u\wedge v$. Its value on the Kähler class is positive: $Q([\omega ],[\omega ])={\int }_M{\omega }^2>0$. Every real harmonic $(1,1)$ form splits uniquely as $c\omega +u_0$, with $u_0$ primitive and $c$ constant. Indeed, $\Lambda u$ is a harmonic function, hence constant by connectedness, and $\Lambda \omega =2$; subtracting $c\omega$ kills $\Lambda u$, and part (b) makes $u_0$ primitive.

Part (c) gives $\ast u_0=-u_0$, while $\omega \wedge u_0=0$. ThusThis proves the Hodge index theorem for compact Kähler surfaces in this case: $Q$ has signature $(1,b_2-1)$.

On the alleged two-dimensional totally isotropic subspace $V$, the linear functional $v\mapsto Q(v,[\omega ])$ has a nonzero vector in its kernel. Its harmonic differential form representative is primitive, so the preceding negative-definiteness statement gives $Q(v,v)<0$, contradicting the vanishing of $Q$ on $V$. ThereforeConnectedness is essential if one's definition of a manifold allows disconnected spaces. For $M=Y\bigsqcup Y$, with $Y={\mathbb{CP}}^1\times {\mathbb{CP}}^1$, one has $H^2(M,{\mathcal{O}}_M)=0$. Take on each component the pullback of the degree-two class from its first factor. Each class has square zero, and their mutual pairing is zero because their supports lie on different components. Their span is then a two-dimensional totally isotropic subspace, disproving the unqualified disconnected version.