MY-ASSIGNMENTEXPERT™可以为您提供math.sjsu.edu MATH213B Riemann Surfaces黎曼曲面课程的代写代考和辅导服务！
MATH213B课程简介
Description
Fundamentals of algebraic curves as complex manifolds of dimension one. Topics may include branched coverings, sheaves and cohomology, potential theory, uniformization and moduli. Course site: https://locator.tlt.harvard.edu/course/colgsas-111824/2022/spring/13552
Credits
4
Recent Professors
Peter Kronheimer, Yum-Tong Siu, Shing-Tung Yau, Curtis McMullen
Prerequisites
In mathematics, particularly in complex analysis, a Riemann surface is a connected one-dimensional complex manifold. These surfaces were first studied by and are named after Bernhard Riemann. Riemann surfaces can be thought of as deformed versions of the complex plane: locally near every point they look like patches of the complex plane, but the global topology can be quite different. For example, they can look like a sphere or a torus or several sheets glued together.
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Recent Semesters
Spring 2023, Spring 2022, Spring 2021, Spring 2020, Spring 2019
Usually Held
TuTh (10:30am-11:45am), TuTh (3:00pm-4:15pm), TuTh (9:00am-10:15am)
MATH213B Riemann Surfaces HELP（EXAM HELP， ONLINE TUTOR）
Let $\omega \in \Omega(X)$ be a holomorphic 1-form on a Riemann surface $X$. Show that locally $\omega=\partial u$, where $u$ is a real-valued harmonic function.
Since $X$ is a Riemann surface, it is a complex manifold of complex dimension 1. Let $(U,z)$ be a local coordinate chart on $X$, where $U$ is an open subset of $X$ and $z$ is a holomorphic coordinate function on $U$. Then $\omega=f(z)dz$ for some holomorphic function $f(z)$ on $U$.
Let $u(z,\bar{z})= \frac{1}{2}(\phi(z)+\phi(\bar{z}))$, where $\phi(z)=\int_{z_0}^{z}f(\zeta)d\zeta$ and $z_0$ is a fixed point in $U$. Note that $\phi(z)$ is well-defined since the integral of a holomorphic function along a path depends only on the endpoints of the path by Cauchy’s theorem. Furthermore, $u(z,\bar{z})$ is a real-valued function on $U$.
We claim that $\omega=\partial u$, where $\partial=\frac{1}{2}(\frac{\partial}{\partial z}+\frac{\partial}{\partial \bar{z}})$ is the Dolbeault operator. To see this, we compute: \begin{align*} \partial u &= \frac{1}{2}(\frac{\partial u}{\partial z}+\frac{\partial u}{\partial \bar{z}}) \ &= \frac{1}{2}(\frac{\partial}{\partial z}\frac{1}{2}(\phi(z)+\phi(\bar{z}))+\frac{\partial}{\partial \bar{z}}\frac{1}{2}(\phi(z)+\phi(\bar{z}))) \ &= \frac{1}{2}(f(z)+0)dz \ &= \omega. \end{align*} The second equality follows from the fact that $u$ is harmonic, i.e., $\frac{\partial^2 u}{\partial z \partial \bar{z}}=0$. Therefore, locally we have $\omega=\partial u$, where $u$ is a real-valued harmonic function.
Show that for any $p \in X$ and $\omega \in \Omega(X)$, there is a local coordinate $z$ at $p$ with $z(p)=0$, in which we have $\omega=z^n d z$ for some $n \geq 0$.
Let $p\in X$ be arbitrary and $\omega \in \Omega(X)$ be given. We can choose a local coordinate $z$ around $p$ such that $z(p) = 0$ and $\omega$ can be expressed as $\omega = f(z) dz$ in this coordinate chart. Since $\omega$ is a holomorphic 1-form, $f(z)$ must be a holomorphic function in a neighborhood of $0$.
Now, we use the power series expansion of $f(z)$ centered at $0$: $$f(z) = \sum_{n=0}^{\infty} a_n z^n,$$ where $a_n = \frac{f^{(n)}(0)}{n!}$ are the Taylor coefficients of $f(z)$ at $0$. Since $f(z)$ is holomorphic at $0$, it follows that $a_n = 0$ for all negative integers $n$. Let $m$ be the smallest nonnegative integer such that $a_m \neq 0$. Then, we have $$\omega = f(z) dz = a_m z^m dz + \sum_{n=m+1}^{\infty} a_n z^n dz = z^m \left(a_m + \sum_{n=m+1}^{\infty} a_n z^{n-m}\right)dz.$$ Define $n = m$ if $a_m \neq 0$, and $n = m+k$ if $a_m = 0$ for $k>0$ and $a_{m+k} \neq 0$. Then, we can write $\omega$ as $\omega = z^n d z$ in the coordinate chart $z$.
Note that $n$ may depend on the choice of the coordinate $z$, but the property that $\omega$ can be written as $z^n d z$ for some $n\geq 0$ is coordinate-independent. This follows from the fact that if we choose a different local coordinate $w$, then $w = g(z)$ for some holomorphic function $g$ in a neighborhood of $0$. Thus, we have $dz = g'(z) dw$ and $z^n d z = z^n g'(z) dw$, so $\omega$ can still be expressed as $z^n d z$ in the coordinate $w$ as well.
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