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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 16: Currents and the Poincar\'e-Dolbeault-Grothendieck lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, March 21, 2018 } \end{center} \newpage {\bf \blue Generalized functions} \definition Let $V$ be a vector space equipped with a collection of norms (or seminorms) $|\cdot|_i$, $i=0, 1, 2, ...$ and a topology which is given by the metric \[ d(x,y) = \sum_{i=0}^\infty 2^{-i}\min(|x-y|_i, 1), \] assumed to be non-degenerate. The space $V$ is called {\bf \blue a Fr\'echet space} if this metric is complete. \remark Completeness {\bf \purple is equivalent to convergence of any sequence $\{a_i\}$ which is fundamental with respect to all the (semi-)norms $|\cdot|_i$.} \remark {\bf \purple A sequence converges in the Fr\'echet topology given by $d$ $\Leftrightarrow$ it converges in any of the (semi-)norms $|\cdot|_i$.} \definition Let $M$ be a Riemannian manifold, and $\nabla^i:\; C^{\infty}(M) \arrow \Lambda^1(M)^{\otimes i}$ the iterated connection. {\bf \blue Topology $C^k$} on the space $C^\infty_c(M)$ of functions with compact support is defined by the norm \[ |\phi|_{C^k}:= \sup_M \sum_{i=0}^k |\nabla^i\phi|.\] \newpage {\bf \blue Generalized functions (2)} \definition {\bf \blue The space of test-functions with compact support} is the space of functions with compact support and a metric \[ d(x,y) = \sum_{i=0}^\infty 2^{-i}\min(|x-y|_{C^i}, 1). \] of uniform convergence of all derivatives. \exercise Prove that {\bf \purple the space of test-functions with support in a compact set $K\subset M$ is a Fr\'echet space.} \definition {\bf \blue Generalized function} (also called {\bf \blue distribution}) is a functional on the space of test-function which is continuous in one of the $C^i$-topologies on the space $C^\infty(M)_{K}$ of functions with support in any compact $K\subset M$. \example {\bf \blue Delta-function} $\delta_z$ is a functional mapping $\phi\in C^\infty_c(M)$ to $\phi(z)$, for a given point $z\in M$. {\bf \purple Delta-function is continuous in the topology $C^0$, its derivative is continuous in $C^1$ and so on}. \newpage {\bf \blue Currents on complex manifolds} \remark The {\bf \blue $C^i$-topology} is defined on the space of sections of any vector bundle $B$ over using the same formula. It depends on the choice of the metric on $M$ and on $B$, but {\bf \red the induced topology is clearly independent from this choice.} \definition {\bf \blue The space of test-forms of type $(p,q)$ on a complex manifold} is the space $\Lambda^{p,q}_c(M)$ with compact support, equipped with the Fr\'echet topology as on the test-functions. \definition A {\bf \blue $(p,q)$-current} on a complex $n$-dimensional manifold is a functional $\theta$ on the space $\Lambda^{n-p,n-q}_c(M)$ of forms with compact support, such that for any compact set $K\subset M$ there exists $i\geq 0$ such that $\theta$ is continuous in $C^i$-topology on forms with support in $K$. \remark {\bf \red A smooth $(p,q)$-form $\psi$ defines a $(p,q)$-current:} given a test-form $\alpha \in \Lambda^{n-p,n-q}_c(M)$, consider the functional $\alpha \arrow \int_M \psi \wedge\alpha$. This gives an embedding $\Lambda^{p,q}(M)\hookrightarrow {\cal D}^{p,q}(M)$ from forms to currents. \remark {\bf \purple Currents are $(p,q)$-forms with coefficients in generalized functions.} \newpage {\bf \blue Cohomology of currents} \definition Define {\bf \blue the de Rham differential on the space of currents} using the formula $ \langle d\psi, \alpha\rangle:= - (-1)^{\tilde \psi}\langle \psi, d\alpha\rangle.$ {\bf \purple This definition is compatible with the embedding $\Lambda^{p,q}(M)\hookrightarrow {\cal D}^{p,q}(M)$ from forms to currents:} \[ \int_M d\psi \wedge \alpha = \int_M d(\psi \wedge \alpha) - (-1)^{\tilde \psi} \int_M \psi \wedge d \alpha = - (-1)^{\tilde \psi} \int_M \psi \wedge d \alpha \] by Stokes' formula. \remark {\bf \blue The Dolbeault differentials} $\6=d^{1,0}$, $\bar\6=d^{0,1}$ are defined on currents using the same formula. \exercise {\bf \red Prove the Poincar\'e lemma for currents.} \definition Let $f:\; X \arrow Y$ be a proper holomorphic map of complex manifolds, $\dim_\C X = \dim_\C Y +k$, and $\alpha$ a $(p,q)$-current on $X$. Define {\bf \blue the pushforward} $f_* \alpha$ using $\langle f_*\alpha, \tau\rangle := \langle \alpha, f^*\tau\rangle$, where $\tau$ is any test-form. Then {\bf \red $f_*\alpha$ has bidimension $(p-k,q-k)$.} One should think of $f_*$ as of fiberwise integration. \remark Clearly, $d f_* \alpha= f_* d\alpha$, $\6 f_* \alpha= f_* \6\alpha$, and so on. \remark Pullback of currents is (generally speaking) not well-defined. \newpage {\bf \blue Poincar\'e-Lelong formula} \claim {\bf \blue (Poincar\'e-Lelong formula)}\\ Consider a current on $\C$ given by $\frac 1{\pi z}d z $. {\bf \red Then $d \left(\frac 1 {\pi z}d z\right ) = \delta_0\Vol$}, where $\delta_0$ is $\delta$-function in 0. \proof For any function smooth $f$ on a closure of a disc $D$ and $w\in D$, Cauchy formula gives \[ f(w) = \frac 1 {2\pi \1} \int_{\6 D}\frac{f(z)}{z-w} dz - \frac 1 \pi \int_D \frac {\bar \6 f}{z-w} \wedge dz. \] Applying this to a test-function $f$ with compact support inside $D$, we obtain \[ f(w)=- \left\langle\frac 1 {\pi z} dz, \bar \6 f\right\rangle = \left\langle\bar \6 \left(\frac 1 {\pi z}\right) d z, f\right\rangle = \left\langle d\left(\frac {d z} {\pi z}\right) , f\right\rangle. \] (the last equality is true because $d\eta=\bar\6\eta$ for any $(1,0)$-form on a disc). \endproof \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck (dimension 1)} \corollary Let $\pi_1, \pi_2:\; \C^2 \arrow \C$ be coordinate projections, and $\xi$ a (1,0)-current on $\C^2$ defined by $\xi:=\frac1{\pi(z-w)}dw$, where $w,z$ are coordinates on $\C^2$. Consider {\bf \blue convolution with the current $\xi$}, given by $P_\xi(\tau):= {\pi_2}_*(\pi^*_1 \tau\wedge \xi)$. {\bf \red Then $\bar\6 P_\xi(\alpha)=\alpha$ for any $(0,1)$-form $\alpha$ with compact support.} \proof $\bar\6 P_\xi(\alpha)= {\pi_2}_*(\pi^*_1 \alpha\wedge \bar\6\xi)= {\pi_2}_*(\pi^*_1 \alpha \wedge \delta_{\triangle})= \alpha,$ where $\delta_{\triangle}$ is $\delta$-function of the diagonal $\triangle$, which is defined as $\langle \kappa, \delta_{\triangle}\rangle := \int_\triangle \kappa$. \endproof \corollary {\bf \red For any (0,1)-form $\alpha$ with compact support on $\C$ there exists a function $f\in C^\infty(\C)$ such that $\bar\6f=\alpha$.} Moreover, $f$ can be chosen in such a way that $|f(z)| < C \frac 1 {|z|}$ for some constant $C>0$ depending on $\int_\C |\alpha|$. \proof Take $f= P_\xi(\alpha)$. From the definition of $P_\xi$ we obtain $|f(z)|< \dist(z, S)^{-1}\int_\C |\alpha|$, where $S=\Supp(\alpha)$. This implies the estimate. \endproof \remark Similarly, for any $(1,1)$-form $\alpha$ with compact support one has $\bar \6(P_\xi(\alpha))=\alpha$, {\bf \purple with the same asymptotic estimates on $P_\xi(\alpha)$.} \end{document}