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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Symplectic geometry\\[15mm] \small lecture 14: K\"ahler reduction} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 20, 2021 } \end{center} \newpage {\bf \blue Characteristic foliation (reminder)} \definition Let $(M, \omega)$ be a symplectic manifold. A submanifold $Z \subset M$ is called {\bf \blue coisotropic} if $\dim_\R Z \geq \dim_\R M$ and $\omega\restrict Z$ has rank $\dim_\R M- \dim_\R Z$ (minimal possible), or, equivalently, $(TZ)^{\bot_\omega} \subset TZ$, where $(TZ)^{\bot_\omega}:= \{ x\in TZ\ \ |\ \ i_x\omega=0\}$ \definition Let $Z\subset (M, \omega)$ be a coisotropic submanifold. The bundle $(TZ)^{\bot_\omega}$ is called {\bf \blue the characteristic bundle of $Z$}. \theorem Let $Z\subset (M, \omega)$ be a coisotropic submanifold, and $K\subset TM$ its characteristic bundle. {\bf \red Then $[K, K]\subset K$, hence $K$ is tangent to a foliation} ${\cal F}$ which is called {\bf \blue the characteristic foliation of $Z$.} Moreover, {\bf \red the restriction $\omega\restrict Z$ is basic, and symplectic on the leaf space of the characteristic foliation.} \newpage {\bf \blue Moment maps (reminder)} \definition $(M,\omega)$ be a symplectic manifold, and $G$ a Lie group acting on $M$ by symplectomorphisms. {\bf \blue A moment map} $\mu$ of this action is a linear map ${\goth g}\arrow C^\infty M$ associating to each $g\in G$ its Hamiltonian. \remark It is more convenient to consider $\mu$ as an element of ${\goth g}^* \otimes_\R C^\infty M$, or (and this is most standard) {\bf \red as a function with values in ${\goth g}^*$}. \remark Moment map {\bf \purple always exists} if $M$ is simply connected. \definition A moment map $M \arrow {\goth g}^*$ is called {\bf \blue equivariant} if it is equivariant with respect to the coadjoint action of $G$ on ${\goth g}^*$. \remark $M\stackrel\mu \arrow {\goth g}^*$ is a moment map iff for all $g\in {\goth g}$, $\langle d\mu,g\rangle= i_{\rho_g}(\omega)$. Therefore, {\bf \purple a moment map is defined up to a constant ${\goth g}^*$-valued function.} An equivariant moment map is is defined up to {\bf \purple a constant ${\goth g}^*$-valued function which is $G$-invariant}, that is, up to addition of a central vector $c\in \g^*$. \claim {\bf \red An equivariant moment map exists whenever $H^1(G, {\goth g}^*)=0$.} In particular, when $G$ is reductive and $M$ is simply connected, an equivariant moment map exists. Further on, all moment maps will be tacitly considered equivariant. \newpage {\bf \blue Weinstein-Marsden theorem (reminder)} %\definition A $G$-invariant $c\in \goth g^*$ is called %{\bf\blue central}. \definition (Weinstein-Marsden) $(M,\omega)$ be a symplectic manifold, $G$ a compact Lie group freely acting on $M$ by symplectomorphisms, $M\stackrel\mu \arrow {\goth g}^*$ an equivariant moment map, and $c\in {\goth g}^*$ a central element. The quotient $\mu^{-1}(c)/G$ is called {\bf \blue symplectic reduction} of $M$, denoted by $M\2 G$. \claim {\bf \red The symplectic quotient $M\2 G$ is a symplectic manifold of dimension $\dim M - 2 \dim G$.}\\ \pstep $T_x (\mu^{-1}(c))= d\mu^{-1}(0)$. However, the space $\langle d\mu, g\rangle \subset \Lambda^1 M$ is $\omega$-dual to the space $\tau(\goth g)$ of vector fields tangent to the $G$-action, {\bf \purple hence $d\mu^{-1}(c)= \tau(\goth g)^\bot$.} {\bf \green Step 2:} Since $\mu$ is $G$-equivariant, $G$ preserves $\mu^{-1}(c)$, hence $\tau(\goth g)\subset d\mu^{-1}(0)$. This implies that {\bf \purple $\tau(\goth g)\subset TM$ is isotropic} (that is, $\omega\restrict{\tau(\goth g)}=0$). Its $\omega$-orthogonal complement in $T_x M$ is $T_x (\mu^{-1}(c))$ (Step 1). {\bf \green Step 3:} Consider the {\bf \blue characteristic foliation} ${\cal F}$ on $\mu^{-1}(c)$. It is a bundle because $\mu^{-1}(c)\subset M$ is coisotropic. From Step 2 {\bf \purple we obtain that ${\cal F}=\tau(\goth g)$.} {\bf \green Step 4:} Since $\omega\restrict {\mu^{-1}(c)}$ is closed, it satisfies $\Lie_v (\omega)=0$ for all $v\in {\cal F}$. This implies that it is {\em \green basic}, that is, lifted from the leaf space of characteristic foliation, identified with $M\2 G$. \endproof \newpage {\bf \blue K\"ahler manifolds} \definition {\bf \blue K\"ahler manifold} is a complex manifold equipped with a compatible symplectic structure. \example {\bf \purple A complex submanifold $Z \subset (M, I, \omega)$ of a K\"ahler manifold is also K\"ahler.} Indeed, a restriction of a symplectic form to $Z$ is closed; it is non-degenerate because $\omega(x, Ix)$ is positive definite, hence $\omega$ is non-degenerate on every $I$-invariant subspace. \example {\bf \purple The Fubini-Studi form $\omega$ on $\C P^n$ is K\"ahler.} \definition A complex submanifold of $\C P^n$ is called {\bf \blue projective}. \example {\bf \purple All projective manifolds are K\"ahler}. \example Compact complex tori {\bf \purple are always K\"ahler, but not always projective.} \newpage {\bf \blue Symplectic reduction and a K\"ahler potential} \definition Let $d^c:= I dI^{-1}$. {\bf \blue K\"ahler potential} on a K\"ahler manifold $(M, \omega)$ is a function $\psi$ such that $dd^c \psi=\omega$. \proposition Let $G$ be a real Lie group acting on a K\"ahler manifold $M$ by holomorphic isometries, and $\psi$ a $G$-invariant K\"ahler potential. {\bf \red Then the moment map $\g \times M \stackrel {\mu_g} \arrow \R$ can be written as $g, m \arrow \Lie_{Iv}\psi$,} where $v=\tau(g)\in TM$ is the tangent vector field associated with $g\in \goth g$. \proof Since $\psi$ is $G$-invariant, and $I$ is $G$-invariant, we have $0 = \Lie_v d^c \psi = i_v (dd^c\psi) + d(i_v d^c \psi)$. Using $\omega=dd^c \psi$, we rewrite this equation as $i_v\omega = -d(\langle d^c \psi, v\rangle)$, giving an equation for the moment map $\mu_g= -\langle d^c \psi, v\rangle.$ Acting by $I$ on both sides, we obtain $\mu_g= \langle d \psi, Iv\rangle = \Lie_{Iv}\psi$. \endproof \corollary Let $V$ be a Hermitian representation of a compact Lie group $G$. {\bf \purple Then the corresponding moment map can be written as $\mu_g(v)= \Lie_{Ig}|v|^2= \frac 1 4 \langle v, Ig(v)\rangle.$} \endproof \newpage {\bf \blue Transversal complex and Riemannian structures} \definition {\bf\blue A foliation} on a manifold $M$ is a sub-bundle $B\subset TM$ such that $[B, B] \subset B$. By Frobenius theorem, this is equivalent to a local decomposition $U= S\times R$, of any sufficiently small open set $U\subset M$, with $B\subset TU$ equal to the tangent bundle to the fibers of the projection $U \arrow R$. {\bf\blue A leaf} of a foliation is a maximal connected immersed submanifold $Z\arrow M$ which satisfies $T_z Z= B\restrict z$ at each $z\in Z$. {\bf\blue Projection to the leaf space} is a smooth submersion $U\arrow R$, mapping $U$ to the set of leaves of $B$ on $U$. \definition {\bf\blue Transversal Riemannian structure / symplectic structure / almost complex structure} on a foliated manifold $(M, B\subset TM)$ is a scalar product / skew-symmetric form/almost complex structure on the bundle $(TM/B)$ which is locally obtained as a pullback of a Riemannian /symplectic / almost complex structure on the leaf space. \newpage {\bf \blue Transversal complex and Riemannian structures (2)} \proposition Let $M$ be a manifold equipped with a locally free action of a compact Lie group $G$ and $B\subset TM$ the bundle of vectors tangent to the $G$-action. Suppose that $TM/B$ is equipped with a $G$-invariant metric $h$ and a $G$-invariant almost complex structure $I$. {\bf \red Then these structures are transversal.} \pstep Let $h\in \Sym^2(T^*M)$ be a symmetric 2-form vanishing on $B$ and positive definite and $G$-invariant on $TM/B$. In Lecture 13 we proved that {\bf \purple a $B$-basic differential form $\alpha$ is the one which satisfies $i_X\alpha=0$ and $\Lie_X \alpha=0$ for all vector fields $X\in B$.} The same argument can be applied to show that $h$ is basic whenever $\Lie_X h=0$ for all $X\in B$. This follows because $h$ is $G$-invariant. {\bf \green Step 2:} It remains to show that the 2-form $\omega\in \Lambda^2 M$ obtained as $\omega(x, y) = h(x, Iy)$ is basic. This form vanishes on $B$ and is $G$-invariant, hence it is basic by the same argument. \endproof \newpage {\bf \blue Symplectic reduction for almost K\"ahler manifolds} %\remark Recall that {\bf \blue an almost K\"ahler manifold} %is a symplectic manifold with compatible almost complex structure. {\bf \green THEOREM 1:} Let $M$ be an almost K\"ahler manifold, $G$ a compact Lie group acting on $M$ freely by Hamiltonian isometries, $t\in \g^*$ a central element, and $\mu:\; M \arrow \g^*$ an equivariant moment map. Denote by $K \subset T\mu^{-1}(t)$ the bundle of vectors tangent to the orbits of $G$ on $\mu^{-1}(t)$. Then $K\subset T\mu^{-1}(t)$ is tangent to a foliation, {\bf \red equipped with a transversal Riemannian and a transversal symplectic structure,} obtained by restricting the Riemannian and symplectic forms to $K^\bot$. Moreover, these bilinear forms {\bf \red define an almost K\"ahler structure on the orbit space $\frac{\mu^{-1}(t)}{G}$.} \proof These two bilinear forms are transversal because they are $G$-invariant, and the symplectic form on $M/G$ is closed by Marsden-Weinstein theorem. \endproof \corollary Let $G$ be a compact group acting by Hamiltonian symplectomorphisms on an almost K\"ahler manifold. Assume that $G$ preserves the almost complex structure. {\bf \red Then the symplectic reduction $M\2 G$ is also almost K\"ahler.} \endproof \remark Since $K\subset T\mu^{-1}(t)$ is isotropic, one has $K \cap I(K)=0$. This implies $T\mu^{-1}(t)/K = TM\restrict{\mu^{-1}(t)}/ K_\C$, where $K_\C = K \oplus I(K)$. {\bf \purple This is another way to equip the bundle $T\mu^{-1}(t)/K$ with an almost complex structure.} \newpage {\bf \blue CR-manifolds} \definition Let $B\subset TM$ be a sub-bundle equipped with a complex structure operator $I\in \End B$, $I^2=-\Id$, and $B\otimes_\R \C= B^{1,0} \oplus B^{0,1}$ the corresponding eigenspace decomposition, $I\restrict{B^{1,0}}=\1$, $I\restrict{B^{0,1}}=-\1$. The pair $(B, I)$ is called {\bf \blue a CR-structure on $M$} if $[B^{1,0}, B^{1,0}] \subset B^{1,0}$. \claim Let $Z\subset M$ be a submanifold of an almost complex manifold. Assume that $B:=TZ \cap I(TZ)$ has constant rank. {\bf \red Then $(B, I \restrict B)$ is a CR-structure on $M$.} \proof Any $(1, 0)$-vector fields $X,Y\in B^{0,1}\subset TZ\otimes_\R \C$ can be smoothly extended to a section $\tilde X, \tilde Y \in T^{1,0}M$. A commutator $[\tilde X,\tilde Y]$ of $(1,0)$-vector fields on $M$ is of type $(1,0)$. Commutator of vector fields tangent to $Z$ remains tangent to $Z$. {\bf \purple Therefore, $[\tilde X,\tilde Y]\bigg|_ Z \in B^{1,0}$, giving $[X,Y]\in B^{1,0}$.} \endproof \newpage {\bf \blue CR-manifolds and leaf spaces} {\bf \green PROPOSITION 2:} Let $(Z, B, I)$ be a CR-manifold, equipped with an action of a compact group $G$ compatible with the CR-structure. Assume that $TZ= B \oplus K$, where $K \subset TZ$ is the subspace generated by the vector fields tangent to the $G$-action. Then the operator $I$ on $B$ {\bf \purple defines a transversal complex structure} with respect to the foliation tangent to $K$. Morever, {\bf \red the natural almost complex structure on $Z/G$ induced by the action of $I$ on $B= TZ/K= T(Z/G)$ is integrable.} \proof This operator is transversal because it is $G$-invariant, and integrable because $[B^{1,0}, B^{1,0}] \subset B^{1,0}$. \endproof \newpage {\bf \blue K\"ahler reduction} \corollary Let $M$ be a K\"ahler manifold, $G$ a compact Lie group freely acting on $M$ by Hamiltonian isometries, $t\in \g^*$ a central element, and $\mu:\; M \arrow \g^*$ an equivariant moment map. {\bf \red Then the quotient $M\2 G:= \frac{\mu^{-1}(t)}{G}$ is equipped with a natural K\"ahler structure.} \pstep Let $Z:= \mu^{-1}(t)$ and let $K\subset TZ$ be the kernel of the restriction of the K\"ahler form to $Z$. {\bf \purple The space $Z:= \mu^{-1}(t)$ is equipped with a CR-structure because $Z$ is a subspace of a complex manifold with $TZ \cap I(TZ)$ of constant rank. } Indeed, $I(TZ) \cap K=0$ because $K$ is isotropic, hence \[ \rk(TZ \cap I(TZ))\leq \dim Z - \rk K= \dim M - 2 \codim Z.\] However, the intersection $TZ \cap I(TZ)$ satisfies $\rk(TZ \cap I(TZ))\geq \dim M - 2 \codim Z$ because it is an intersection of two subspaces of codimension $\codim Z$, giving $\rk(TZ \cap I(TZ))=\dim Z - \rk K$. {\bf \green Step 2:} The quotient $M\2 G:= Z/G$ is almost K\"ahler by Theorem 1. The almost complex structure on $M\2 G$ is integrable by Proposition 2. \endproof \end{document}