Solutions to exercises

Exercise 3.3

Show that, for ${\displaystyle \omega }$ a Hermitian form, ${\displaystyle \omega (X-iJX,Y-iJY)=0}$.

Proof

Let us start by just using the definition of ${\displaystyle \omega }$ in terms of our Riemannian metric ${\displaystyle g:}$

{\displaystyle {\begin{aligned}\omega (X-iJX,Y-iJY)&=g(J(X-iJX),Y-iJY)\\&=g(JX+iX,Y-iJY)\end{aligned}}}
Here we've used ${\displaystyle J^{2}=-\mathrm {Id} }$. Now using bilinearity of ${\displaystyle g}$ we get
${\displaystyle g(JX+iX,Y-iJY)=g(JX,Y)+ig(X,Y)-ig(JX,JY)+g(X,JY)}$
Now we recall that ${\displaystyle g(JX,JY)=g(X,Y)}$, so we obtain
${\displaystyle g(-X,JY)++ig(X,Y)-ig(X,Y)+g(X,JY)=0}$
as required.

Exercise 3.4

Prove that any projective manifold admits a Kähler metric.

Proof

The solution to this exercise boils down to the construction of the Fubini-Study metric on ${\displaystyle \mathbb {CP} ^{n}}$. Indeed, suppose we are provided with a Kähler metric on the projective space ${\displaystyle \mathbb {CP} ^{n}}$ and let ${\displaystyle X\subset \mathbb {CP} ^{n}}$ be a complex (projective) submanifold and let us denote with ${\displaystyle \iota }$ the inclusion ${\displaystyle \iota :X\rightarrow \mathbb {CP} ^{n}}$. We claim that, if ${\displaystyle \omega }$ is a Kähler form on ${\displaystyle \mathbb {CP} ^{n}}$, then ${\displaystyle \omega _{X}:=\iota ^{*}\omega }$ is a Kähler form on ${\displaystyle X}$: in fact, ${\displaystyle \omega _{X}}$ is a real ${\displaystyle (1,1)}$-form, ${\displaystyle \omega _{X}(\cdot ,J_{X}\cdot )}$ is positive defined and ${\displaystyle d\omega _{X}=0}$, because the pullback commutes with the exterior differential and ${\displaystyle \omega }$ is closed. Given this, let us show the construction of the Fubini-Study metric on ${\displaystyle \mathbb {CP} ^{n}}$: consider local sections ${\displaystyle s_{1},s_{2}}$ of the tautological bundle ${\displaystyle {\mathcal {O}}_{\mathbb {CP} ^{n}}(-1)\rightarrow \mathbb {CP} ^{n}}$, ${\displaystyle s_{i}:U\rightarrow {\mathcal {O}}_{\mathbb {CP} ^{n}}(-1)}$, which can be thought as a function ${\displaystyle s_{i}:U\rightarrow \mathbb {C} ^{n+1}}$. Then, by definition of line bundle, there is a holomorphic function ${\displaystyle \lambda :U\rightarrow \mathbb {C} ^{*}}$ such that ${\displaystyle s_{1}=\lambda s_{2}}$. Therefore, ${\displaystyle ||s_{1}||^{2}=||\lambda ||^{2}||s_{2}||^{2}}$ and this fact proves that the quatity ${\displaystyle \partial {\bar {\partial }}\log ||s||^{2}}$ does not depend on the section we choose. We thus can define a global real ${\displaystyle (1,1)}$-form ${\displaystyle \omega _{FS}}$ by setting

${\displaystyle \omega _{FS}:={\frac {i}{2\pi }}\partial {\bar {\partial }}\log ||s||^{2},}$
where ${\displaystyle s}$ is a local section of the tautological bundle. Using the identities ${\displaystyle \partial ^{2}={\bar {\partial }}^{2}=\partial {\bar {\partial }}+{\bar {\partial }}\partial =0}$, one can easily check that ${\displaystyle \omega _{FS}}$ is closed. In order to see that it is the form associated to a riemannian metric, we write it in local coordinates, choosing ${\displaystyle s}$ to be ${\displaystyle s([Z_{0},\dots ,Z_{n}])=(Z_{0},\dots ,Z_{n})}$. A straightforward computation shows that, at ${\displaystyle [1,0,\dots ,0]\in U_{0}=\{z_{0}\neq 0\}}$,
${\displaystyle \omega _{FS}(0,\dots ,0)={\frac {i}{2\pi }}\sum _{i}dz_{i}\wedge d{\bar {z}}_{i}}$
and the corresponding Hermitian metric is a real scalar multiple of the identity, which is positive definite. We claim that it is enough to check the positive definiteness of the form only at one point: indeed ${\displaystyle \omega _{FS}}$ is ${\displaystyle U(n+1)}$-invariant, and ${\displaystyle U(n+1)}$ acts trasitively on ${\displaystyle \mathbb {CP} ^{n}}$. This concludes our proof.

Exercise 3.5

Show that 4k dimensional spheres do not admit almost complex structures.

Proof

The basic topological feature of the tangent bundle is that it is stably trivial. This is because the sphere sits inside the Euclidean space, and the normal bundle is a 1D trivial bundle. Thus ${\displaystyle \mathbb {R} ^{4k+1}=\mathbb {R} \bigoplus TS^{4k}}$.

This means the Chern classes of the complexified tangent bundle are trivial.

Now suppose we have an almost complex structure, then the complexified bundle splits into (1,0) and (0,1) parts which are complex vector bundles. Most of the Chern classes vanish for purely degree reasons. But the top Chern class of the (1,0) bundle is the Euler characteristic times the fundamental class, by Gauss Bonnet. Now since the top Chern class is ${\displaystyle c_{2k}}$ we observe the conjugate bundle has the same Chern class. Then an application of Whitney sum formula gives a contradiction of the value of the top Chern class of the complexified tangent bundle.