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Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Hodge theory \\[15mm] \small lecture 11: K\"ahler manifolds} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, February 28, 2018 } \end{center} \newpage {\bf \blue Almost complex manifolds (reminder)} \definition Let $I:\; TM \arrow TM$ be an endomorphism of a tangent bundle satisfying $I^2=-\Id$. Then $I$ is called {\bf \blue almost complex structure operator}, and the pair $(M,I)$ {\bf \blue an almost complex manifold}. \example $M=\C^n$, with complex coordinates $z_i=x_i + \1 y_i$, and $I(d/dx_i)=d/dy_i$, $I(d/dy_i)=-d/dx_i$. \definition Let $(V,I)$ be a space equipped with a complex structure $I:\; V \arrow V$, $I^2 =-\Id$. {\bf \blue The Hodge decomposition} $V\otimes_\R \C:= V^{1,0}\oplus V^{0,1}$ is defined in such a way that $V^{1,0}$ is a $\1$-eigenspace of $I$, and $V^{0,1}$ a $-\1$-eigenspace. \definition A function $f:\; M \arrow \C$ on an almost complex manifold is called {\bf \blue holomorphic} if $df \in \Lambda^{1,0}(M)$. \remark For some almost complex manifolds, {\bf \red there are no holomorphic functions at all}, even locally. %Example: $S^6$ with a certain %canonical ($G_2$-invariant) almost complex structure. \newpage {\bf \blue Complex manifolds and almost complex manifolds (reminder)} \definition {\bf \blue Standard almost complex structure} is $I(d/dx_i)=d/dy_i$, $I(d/dy_i)=-d/dx_i$ on $\C^n$ with complex coordinates $z_i=x_i+\1 y_i$. \definition A map $\Psi:\; (M,I)\arrow (N,J)$ from an almost complex manifold to an almost complex manifold is called {\bf \blue holomorphic} if $\Psi^*(\Lambda^{1,0}(N))\subset \Lambda^{1,0}(M)$. \remark This is the same as $d\Psi$ being complex linear; for standard almost complex structures, {\bf \purple this is the same as the coordinate components of $\Psi$ being holomorphic functions.} \definition {\bf \blue A complex manifold} is a manifold equipped with an atlas with charts identified with open subsets of $\C^n$ and transition functions holomorphic. \newpage {\bf \blue Integrability of almost complex structures (reminder)} \definition An almost complex structure $I$ on a manifold is called {\bf\blue integrable} if any point of $M$ has a neighbourhood $U$ diffeomorphic to an open subset of $\C^n$, in such a way that the almost complex structure $I$ is induced by the standard one on $U\subset \C^n$. \claim {\bf \purple Complex structure on a manifold $M$ uniquely determines an integrable almost complex structure, and is determined by it.} \proof Complex structure on a manifold $M$ is determined by the sheaf of holomorphic functions $\calo_M$, because $d\calo_M$ generates $\Lambda^{1,0}(M)$, and $\calo_M$ is determined by $I$, because $\calo_M=\{f\ \ |\ \ df\in \Lambda^{1,0}(M)\}$. \endproof \newpage {\bf \blue Formal integrability (reminder)} \definition An almost complex structure $I$ on $(M,I)$ is called {\bf\blue formally integrable} if $[T^{1,0}M, T^{1,0}]\subset T^{1,0}$, that is, if $T^{1,0}M$ is involutive. \definition The Frobenius form $\Psi\in \Lambda^{2,0}M\otimes TM$ is called {\bf \blue the Nijenhuis tensor}. \claim {\bf \purple If a complex structure $I$ on $M$ is integrable, it is formally integrable.} \proof Locally, the bundle $T^{1,0}(M)$ is generated by $d/dz_i$, where $z_i$ are complex coordinates. These vector fields commute, hence satisfy $[d/dz_i, d/dz_j]\in T^{1,0}(M)$. This means that the Frobenius form vanishes. \endproof \theorem {\bf \blue (Newlander-Nirenberg)}\\ {\bf\red A complex structure $I$ on $M$ is integrable if and only if it is formally integrable.} \remark {\bf \purple In dimension 1, formal integrability is automatic.} Indeed, $T^{1,0}M$ is 1-dimensional, hence all skew-symmetric 2-forms on $T^{1,0}M$ vanish. \newpage {\bf \blue Riemannian manifolds} \definition Let $h\in \Sym^2 T^* M$ be a symmetric 2-form on a manifold which satisfies $h(x,x)>0$ for any non-zero tangent vector $x$. Then $h$ is called {\bf \blue Riemannian metric}, of {\bf \blue Riemannian structure}, and $(M,h)$ {\bf \blue Riemannian manifold}. \definition For any $x, y\in M$, and any path $\gamma:\; [a, b] \arrow M$ connecting $x$ and $y$, consider {\bf \blue the length} of $\gamma$ defined as $L(\gamma) = \int_\gamma |\frac{d\gamma}{dt}| dt$, where $|\frac{d\gamma}{dt}|= h(\frac{d\gamma}{dt},\frac{d\gamma}{dt})^{1/2}$. Define {\bf \blue the geodesic distance} as $d(x, y) = \inf_\gamma L(\gamma)$, where infimum is taken for all paths connecting $x$ and $y$. \exercise Prove that the {\bf \purple geodesic distance satisfies triangle inequality and defines metric on $M$.} \exercise Prove that {\bf \purple this metric induces the standard topology on $M$.} \example Let $M=\R^n$, $h= \sum_i dx_i^2$. {\bf \purple Prove that the geodesic distance coincides with $d(x, y)= |x-y|$.} \exercise Using partition of unity, {\bf \red prove that any manifold admits a Riemannian structure.} \newpage {\bf \blue K\"ahler manifolds} {\bf\green DEFINITION:} An Riemannian metric $g$ on an almost complex manifiold $M$ is called {\bf \blue Hermitian} if $g(Ix, Iy)= g(x,y)$. In this case, $g(x, Iy)= g(Ix, I^2y) = - g(y, Ix)$, hence $\omega(x,y):= g(x, Iy)$ is skew-symmetric. \remark Given any Riemannian metric $g$ on an almost complex manifold, {\bf \purple a Hermitian metric $h$ can be obtained as $h= g + I(g)$, where $I(g)(x,y)=g(I(x),I(y))$.} {\bf\green DEFINITION:} The differential form $\omega\in \Lambda^{1,1}(M)$ is called {\bf \blue the Hermitian form} of $(M,I,g)$. \remark It is $U(1)$-invariant, hence {\bf \purple of Hodge type (1,1)}. \remark In the triple $I, g, \omega$, {\bf \purple each element can recovered from the other two.} {\bf\green DEFINITION:} A complex Hermitian manifold $(M,I,\omega)$ is called {\bf \blue K\"ahler} if $d\omega=0$. The cohomology class $[\omega]\in H^2(M)$ of a form $\omega$ is called {\bf \blue the K\"ahler class} of $M$, and $\omega$ {\bf \blue the K\"ahler form}. \newpage {\bf \blue Homogeneous spaces} \definition {\bf \blue A Lie group} is a smooth manifold equipped with a group structure such that the group operations are smooth. Lie group $G$ {\bf \blue acts on a manifold $M$} if the group action is given by the smooth map $G \times M \arrow M$. \definition Let $G$ be a Lie group acting on a manifold $M$ transitively. Then $M$ is called {\bf \blue a homogeneous space}. For any $x\in M$ the subgroup $\St_x(G)=\{g\in G\ \ |\ \ g(x)=x\}$ is called {\bf \blue stabilizer of a point $x$}, or {\bf \blue isotropy subgroup}. \claim For any homogeneous manifold $M$ with transitive action of $G$, {\bf \purple one has $M=G/H$,} where $H=\St_x(G)$ is an isotropy subgroup. \proof The natural surjective map $G\arrow M$ putting $g$ to $g(x)$ identifies $M$ with the space of conjugacy classes $G/H$. \endproof \remark Let $g(x)=y$. Then $\St_x(G)^g=\St_y(G)$: {\bf \purple all the isotropy groups are conjugate.} \newpage {\bf \blue Isotropy representation} \definition Let $M=G/H$ be a homogeneous space, $x\in M$ and $\St_x(G)$ the corresponding stabilizer group. The {\bf \blue isotropy representation} is the natural action of $\St_x(G)$ on $T_x M$. \definition A tensor $\Phi$ on a homogeneous manifold $M=G/H$ is called {\bf \blue invariant} if it is mapped to itself by all diffeomorphisms which come from $g\in G$. \remark Let $\Phi_x$ be an isotropy invariant tensor on $\St_x(G)$. For any $y\in M$ obtained as $y=g(x)$, consider the tensor $\Phi_y$ on $T_y M$ obtained as $\Phi_y:=g(\Phi)$. The choice of $g$ is not unique, however, for another $g'\in G$ which satisfies $g'(x)=y$, we have $g=g'h$ where $h\in \St_x(G)$. Since $\Phi$ is $h$-invariant, {\bf \purple the tensor $\Phi_y$ is independent from the choice of $g$.} We proved {\bf \green Theorem 1:} {\bf \red $G$-invariant tensors on $M=G/H$ are in bijective correspondence with isotropy invariant tensors on $T_x M$,} for any $x\in M$. \endproof \newpage {\bf \blue Representations acting transitively on a sphere} \theorem Let $G$ be a group acting on a vector space $V$. Suppose that $G$ acts transitively on a unit sphere $\{ x\in V \ \ |\ \ g(x)=1\}$. {\bf \red Then a $G$-invariant bilinear symmetric form is unique up to a constant multiplier.} \pstep Since $G$ preserves the sphere, which is a level set of the quadratic form $g$, $g$ is $G$-invariant. {\bf \green Step 2:} For any $G$-invariant quadratic form $g'$, the function $x \arrow \frac {g'(x)}{g(x)}$ is constant on spheres and invariant under homothety, hence it is constant. \endproof \exercise Let $V$ be a representation of $G$, and suppose $G$ acts transitively on a sphere. {\bf \purple Prove that $V$ is an irreducible representation.} \exercise Prove the {\bf \blue Schur lemma:} let $V$ be an irreducible representation of $G$ over $\R$, and $g$ a $G$-invariant positive definite bilinear symmetric form. {\bf \red Then any $G$-invariant bilinear symmetric form is proportional to $g$.} \newpage {\bf \blue Fubini-Study form} \example Consider the natural action of the unitary group $U(n+1)$ on $\C P^n$. The stabilizer of a point is $U(n)\times U(1)$. \theorem There exists an $U(n+1)$-invariant Riemann form on $\C P^n$. Moreover, {\red \bf such a form is unique up to a constant multiplier, and K\"ahler.} \remark This Riemannian structure is called {\bf \blue the Fubini-Study metric}, and its Hermitian form {\bf \blue the Fubini-Study form}. \pstep To construct a $U(n+1)$-invariant Riemann form on $\C P^n$, we take a $U(n)$-invariant form on $T_x \C P^n$ and apply Theorem 1. A $U(n)$-invariant form on $T_x \C P^n$ exists, because it is a standard representation. {\bf \green Step 2:} Uniqueness follows because the isotropy group acts transitively on a sphere. \endproof \claim {\bf \red The Fubini-Study form is closed,} and the corresponding metric is K\"ahler. \proof Let $\omega$ be a Fubini-Study form. Then $d\omega$ is an isotropy-invariant 3-form on $T_x \C P^n$. However, the isotropy group contains $-\Id$, {\bf \red hence all isotropy-invariant odd tensors vanish.} \endproof \newpage {\bf \blue Projective manifolds} \definition Let $M$ be a complex manifold, and $X\subset M$ a smooth submanifold. It is called {\bf \blue a complex submanifold} if $I(TX)\subset TX$, and the map $X \hookrightarrow M$ {\bf \blue a complex embedding}. A complex manifold which admits a complex embedding to $\C P^n$ is called {\bf \blue a projective manifold}. \remark {\bf \purple A complex submanifold of a K\"ahler manifold is K\"ahler.} Indeed, restriction of a Hermitian metric is Hermitian, and restriction of a closed form is closed. Therefore, {\bf \red all projective manifolds are K\"ahler}. \definition A subvariety of $\C P^n$ is called {\bf \blue complex algebraic} if can be obtained as common zeroes of a system of homogeneous polynomial equations. \theorem {\bf \blue (Chow theorem)} {\bf \red All complex submanifolds in $\C P^n$ are complex algebraic.} \newpage {\bf \blue Kodaira embedding theorem} \definition {\bf \blue K\"ahler class} of a K\"ahler manifold is the cohomology class $[\omega]\in H^2(M, \R)$ of its K\"ahler form. We say that $M$ {\bf \blue has integer K\"ahler class} if $[\omega]$ belongs to the image of $H^2(M, \Z)$ in $H^2(M, \R)$ \remark $H^2(\C P^n, \R)=\R$. This implies that {\bf \red the cohomology class of Fubini-Study form can be chosen integer.} In particular, {\bf \red all projective manifolds admit K\"ahler structures with integer K\"ahler classes.} \theorem {\bf \blue (Kodaira embedding theorem)} Let $M$ be a K\"ahler manifold with an integer K\"ahler class. {\bf \red Then it is projective.} {\bf \green This theorem will be proven later in these lectures.} \newpage {\bf \blue Classes of almost complex manifolds} \vfill \centerline{\epsfig{file=classes-mflds.eps,width=0.6\linewidth}} \vfill \end{document}