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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 7: K\"ahler metrics on homogeneous spaces} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 14, 2020 } \end{center} \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 bilinear symmetric form (or any 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 $T_x M$, where $M=G/H$ is a homogeneous space. For any $y\in M$ obtained as $y=g(x)$, consider the form $\Phi_y$ 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 \theorem Let $M=G/H$ be a homogeneous space and $x\in M$ a point. Then $G$-invariant tensors on $M=G/H$ {\bf \red are in bijective correspondence with isotropy invariant tensors} on the vector space $T_x M$. \endproof \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 Erich K\"ahler} \begin{center} \epsfig{file=Erich_Kahler-1990.jpeg,height=0.40\linewidth} {\small (Erich K\"ahler: 1990)} {\bf \green \small 16 January 1906 - 31 May 2000} \end{center} \newpage {\bf \blue Chez les Weil. Andr\'e et Simone} Andr\'e Weil: 6 May 1906 - 6 August 1998. \begin{center}\epsfig{file=Simone-et-Andre-Weil-enfants-a-Penthievre.jpg, width=0.30\linewidth} {\it\green ``Simone et Andr\'e \`a Penthi\'evre, 1918-1919''} \end{center} \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 compact 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}