"Compact Kähler manifolds"의 두 판 사이의 차이

2013년 6월 3일 (월) 14:06 판

introduction

Hermitian complex manifold $M$ equipped with a closed Kähler from $\omega$, i.e., $d\omega=0$
$\omega=-2ih_{\alpha\overline{\beta}}dz^{\alpha}dz^{\overline{\beta}}$

dimension 1 case

$h_{\alpha\overline{\alpha}}=h_{\overline{\alpha}\alpha}:=h$
$\omega=-2ih\,dz d\overline{z}$
for $\mathbb{P}^{1}$,

$$ \omega=\frac{-i}{2\pi}\frac{dz d\bar{z}}{(1+|z|^2)^2} $$ see Chern class

examples

cohomology theory

compact Kähler manifold of dimension n
Dolbeault cohomology
$h^{p,q}=\operatorname{dim} H^{p,q}(X)$
$h^{p,q}=h^{q,p}$
Serre duality $h^{p,q}=h^{n-p,n-q}$

Hodge decomposition theorem

Let $M$ be a compact Kähler manifold. Let $H^{p,q}(M)$ be the space of cohomology classes represented by a closed form of type $(p,q)$. There is a direct sum decomposition

$$ H^{m}_{dR}(M;\mathbb{C})=\bigoplus_{p+q=m}H^{p,q}(M) $$ Moreover, $H^{p,q}(M)=\overline{H^{q,p}(M)}$. In other words, $H^{m}_{dR}(M)$ carries a real Hodge structure of weight $m$.

@@ 1번째 줄: / 1번째 줄: @@
+==introduction==
+* Hermitian complex manifold $M$ equipped with a closed Kähler from $\omega$, i.e., $d\omega=0$
+* $\omega=-2ih_{\alpha\overline{\beta}}dz^{\alpha}dz^{\overline{\beta}}$
+==dimension 1 case==
+* $h_{\alpha\overline{\alpha}}=h_{\overline{\alpha}\alpha}:=h$
+* $\omega=-2ih\,dz d\overline{z}$
+* for $\mathbb{P}^{1}$,
+$$
+\omega=\frac{-i}{2\pi}\frac{dz d\bar{z}}{(1+|z|^2)^2}
+$$
+see [[Chern class]]
 ==examples==
 * [[Chern class|complex projective line]]