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\begin{document}
\title{MAT 545 Fall 2025\\ Homework 2}
\maketitle
\begin{itemize}

\item[{\bf 1.}] 
\begin{itemize}
\item[(i)] Let $X$ be a smooth (or complex) manifold with the open covering $\{U_{\alpha}\}_{\alpha\in A}$ and the assignment for each nonempty ordered intersection $U_{\alpha}\cap U_{\beta}$ a smooth (holomorphic) map $g_{\alpha\beta}: U_{\alpha}\cap U_{\beta}\to\mathrm{GL}(r,\mathbb{R})$ (or $\mathrm{GL}(r,\mathbb{C})$), satisfying 
$$g_{\alpha\beta}\cdot g_{\beta\gamma}\cdot g_{\gamma\alpha}=I\quad\text{on}\quad U_{\alpha}\cap U_{\beta}\cap U_{\gamma}\neq\emptyset, \quad\text{and}\quad g_{\alpha\alpha}=I,$$
where $I$ is the identity operator on $\mathbb{R}^{r}$ (or $\mathbb{C}^{r}$). Let
$E$ be smooth (or holomorphic) rank $r$ real (or complex) vector bundle over $X$, defined by
$$E=\bigsqcup_{\alpha\in A}U_{\alpha}\times\mathbb{K}^{r}/\sim\quad (\mathbb{K}=\mathbb{R}\;\text{or}\; \mathbb{C}),$$
where $(x,u)\in U_{\alpha}\times\mathbb{K}^{r}$ is equivalent to $(y,v)\in U_{\beta}\times\mathbb{K}^{r}$ iff $x=y$ and $u=g_{\alpha\beta}(x)v$. Prove that transition functions for $E$ are $g_{\alpha\beta}$.

\item[(ii)] Let $X$ be connected complex manifold and $\mathcal{E}$ be a locally free sheaf of $\mathcal{O}_{X}$-modules over $X$. Prove that there is a holomorphic vector bundle over $X$ whose sheaf of holomorphic sections is $\mathcal{E}$.

\end{itemize}

\vspace{2mm}


\item[{\bf 2.}] Let $X$ be a complex manifold  with the holomorphic atlas $\{(U_{\alpha}, \varphi_{\alpha})\}_{\alpha\in A}$,  $\dim_{\mathbb{C}}X=n$. Show that the functions $g_{\alpha\beta}=J(\varphi_{\alpha}\circ\varphi_{\beta}^{-1})\circ\varphi_{\beta}$, where $J(f)$ is the Jacobian matrix of a holomorphic map $f$ of a domain $D$ in $\mathbb{C}^{n}$ to $\mathbb{C}^{n}$, are transition functions of the holomorphic tangent bundle $\mathcal{T}_{X}$ of $X$. 

\vspace{2mm}
\item[{\bf 3.}] Show that  the complex manifold is canonically oriented.

\vspace{2mm}

\item[\bf{4.}]  Let $D\subset\mathbb{C}^{n}$ be a polydisk, and suppose that $\omega\in\mathcal{A}^{1,1}(D)$ satisfies
$d\omega = 0$ and is real-valued. Prove that there is a smooth function $f:D\to\mathbb{R}$ with the property that $\omega=i\delb\del f$.


\vspace{2mm}


\item[\bf{5.}] Consider  the Bochner-Martinelli kernel
$$k_{\text{BM}}(z)=c_{n}\sum_{k=1}^{n}(-1)^{k-1}\frac{\z_{k}}{|z|^{2n}}dz_{1}\wedge\cdots\wedge dz_{n}\wedge d\z_{1}\wedge\cdots\widehat{d\z_{k}}\wedge\cdots\wedge d\z_{n},$$
where $c_{n}=(-1)^{\frac{n(n-1)}{2}}(n-1)!/(2\pi i)^{n}$, and $\widehat{d\z_{k}}$ means that the factor $d\z_{k}$ is omitted.
\begin{itemize}
\item[(a)] Prove that $d k_{\text{BM}}=\delb k_{\text{BM}}=0$ on $\CC^{n}\setminus\{0\}$.
\item[(b)] Let $S^{2n-1}\subset\RR^{2n}\simeq\CC^{n}$ be the $(2n-1)$-dimensional unit sphere. Show that
$$\int_{S^{2n-1}}k_{\text{BM}}=1.$$
\item[(c)] Let $\pi:\CC^{n}\times\CC^{n}\to\CC^{n}$ be the map $\pi(z,\zeta)=z-\zeta$, and let
$$K(z,\zeta)=\Pi^{0,0}_{z}\Pi^{n,n-1}_{\zeta}\pi^{*} k_{\text{BM}}$$
(note that projector $\Pi^{0,0}_{z}$ is actually redundant since $ k_{\text{BM}}$ has type $(n,n-1)$). Let $D\subset\CC^{n}$ be a bounded domain with a $C^{1}$ boundary and let $f\in C^{1}(\bar{D})$.
 Prove Bochner-Martinelli formula:
 $$f(z)=\int_{\del D}K(z,\zeta)f(\zeta)+\int_{D}K(z,\zeta)\wedge\del f(\zeta).$$
\end{itemize}
\item[\bf{6.}] Let 
$$K^{p,q}(z,\zeta)=\Pi^{p,q}_{z}\Pi^{n-p,n-q-1}_{\zeta}\pi^{*} k_{\text{BM}}$$
(the projector $\Pi^{p,q}_{z}$ is actually redundant), and let $\vp\in\mathcal{A}^{p,q}_{c}(\CC^{n})$ be a differential form of type $(p,q)$ with compact support. Prove Koppelman formula:
$$\vp(z)=\delb_{z}\int_{\CC^{n}}K^{p,q-1}(z,\zeta)\wedge \vp(\zeta)+\int_{\CC^{n}}K^{p,q}(z,\zeta)\wedge\delb\vp(\zeta).$$

\vspace{1mm}
(\emph{Hint:} Use relation $\delb_{\zeta}K^{p,q}=-\delb_{z}K^{p,q-1}$, which is derived from part (a) of problem 5).

\vspace{2mm}
\item[\bf{7.}] Let $B=\{z\in\CC^{n}: |z|<1\}$ be the unit ball in $\CC^{n}$ centered at $0$, $S^{2n-1}=\del B$ and let $f\in\mathcal{O}(\bar{B})$. Using the following steps, 
prove Leray-Fantappi\`{e} formula
$$f(z)=\int_{S^{2n-1}}f(\zeta)S(z,\zeta),$$
where
$$S(z,\zeta)= \frac{(n-1)!}{(2\pi i)^{n}}\sum_{k=1}^{n}(-1)^{k-1}\frac{\bar\zeta_{k}}{(1-\bar\zeta\cdot z)^{n}} d\bar\zeta_{1}\wedge\cdots\widehat{d\bar\zeta_{k}}\wedge\cdots\wedge d\bar\zeta_{n}\wedge d\zeta_{1}\wedge\cdots\wedge d\zeta_{n}$$
and $\bar\zeta\cdot z=\bar\zeta_{1}z_{1}+\cdots+\bar\zeta_{n}z_{n}$.
\begin{itemize}
\item[(a)] Put (as in class) 
\begin{gather*}
\mathcal{K}(z,w,\zeta)=c_{n}\sum_{k=1}^{n}(-1)^{k-1}\frac{(w_{k}-\bar\zeta_{k})}{((z-\zeta)\cdot(w-\bar\zeta))^{n}}
(dz_{1}-d\zeta_{1})\wedge\cdots\wedge (dz_{n}-d\zeta_{n})\wedge\\
(dw_{1}-d\bar\zeta_{1})\wedge\cdots\wedge(dw_{k-1}-d\bar\zeta_{k-1})\wedge(dw_{k+1}-d\bar\zeta_{k+1})\wedge\cdots\wedge(dw_{n}-d\bar\zeta_{n})
\end{gather*}
and
$$\mathcal{K}_{0}(z,w,\zeta)=\Pi^{0,0}_{z}\Pi^{0,0}_{w}\Pi^{n,n-1}_{\zeta}\mathcal{K}(z,w,\zeta),$$
so that 
$$K(z,\zeta)=\mathcal{K}_{0}(z,\z,\zeta)$$
--- a pullback of $\mathcal{K}_{0}$ under the map $\pi:\C^{n}\times\C^{n}\to\C^{n}\times\C^{n}\times\C^{n}$ given by $(z,\zeta)\mapsto (z,\z,\zeta)$.
For $f\in\mathcal{O}(\bar{B})$ and $z,w\in B$ put
$$g(z,w)=\int_{\del B}\mathcal{K}_{0}(z,w,\zeta)f(\zeta).$$
Show that $g\in\mathcal{O}(B\times B)$.
\item[(b)] Using Bochner-Martinelli formula, for $f\in\mathcal{O}(\bar{B})$ and $z\in B$ show that $f(z)=\left.g(z,w)\right|_{w=\z}$.
\item[(c)] Prove that $\del_{w} g(z,w)=0$, so $g(z,w)$ does not actually depend on $w$. Get Leray-Fantappi\`{e} formula by putting $w=0$ in the integral formula in part (a). 
\end{itemize}
\item[\bf{8.}]
Prove that $f\in\mathcal{O}(B)$ admits a power series expansion in $B$, which converges absolutely and uniformly for $z$ in every ball centered at $0$ of radius $r<1$.
\end{itemize}
\end{document}  