Ngô Quốc Anh

December 17, 2009

R-G: Defining function

Filed under: Riemannian geometry — Ngô Quốc Anh @ 15:58

Let $M$ be a smooth manifold of dimension $n$.

Definition. If $S \subset M$ is an embedded submanifold, a smooth map $\Psi : M \to N$ such that $S$ is a regular level set of $\Psi$ is called a defining map for $S$. In other words,

$\displaystyle S=\Psi^{-1}(c)$

for some point $c \in N$. In particular, if $N=\mathbb R^{n-p}$ (so that $\Psi$ is a real-valued or vector-valued function), it is usually called a defining function.

Example 1. The sphere $\mathbb S^n$ is an embeded submanifold of $\mathbb R^{n+1}$. The sphere is easily seen to be a regular level set of the function $f:\mathbb R^{n+1} \to \mathbb R$ given by $f(x)=|x|^2$ since $df=2 \sum_i x^i dx^i$ vanishes only at the origin.

Definition. More generally, if $U$ is an open subset of $M$ and $\Psi:U \to N$ is a smooth map such that $S \cap U$ is a regular level set of $\Psi$, then $\Psi$ is called a local defining map (or local defining function) for $S$.

Example 2. The smooth map $X:\mathbb R^2 \to \mathbb R^3$ given by

$\displaystyle X(\varphi, \theta)=\big( (2+\cos \varphi) \cos \theta, (2+\cos \varphi) \sin \theta, \sin \varphi\big)$

is an immersion of $\mathbb R^2$ into $\mathbb R^3$ whose image, denoted by $D$, is the doughnut-shaped shape surface obtained by revolving the circle $(y-2)^2+z^2=1$ around the $z$-axis, a point $(x,y,z)$ is in $D$ if and only if it satisfies $(r-2)^2+z^2=1$ where $r=\sqrt{x^2+y^2}$ is the distance from the $z$-axis. Thus $D$ is the zero set of the function $\Psi(x,y,z)=(r-2)^2+z^2-1$ which is smooth on $\mathbb R^3$ minus the $z$-axis. A straightforward computation shows that $lated d\Psi$ does not vanish on $D$, so $\Psi$ is a global defining function for $D$.

Schwarz’s Lemma, Schwarz-Pick theorem, and some applications involving inequalities

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 10:52

In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

Schwarz’s Lemma: Let $D=\{z : |z|<1\}$ be the open unit disk in the complex plane $\mathbb C$. Let $f : D \to \overline D$ be a holomorphic function with $f(0)=0$. The Schwarz lemma states that under these circumstances $|f(z)| \leq |z|$ for all $z \in D$, and $|f'(0)| \leq 1$. Moreover, if the equality $|f(z)|=|z|$ holds for any $z \ne 0$, or $|f'(0)|=1$ then $f$ is a rotation, that is, $f(z)=az$ with $|a=1$.

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the “rigidity” of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function $\frac{f(z)}{z}$.

Proof: Let $g(z)=\frac{f(z)}{z}$. The function $g(z)$ is holomorphic in $D$ (excluding $0$) since $f(0)=0$ and $f$ is holomorphic. Let $D_r$ be a closed disc within $D$ with radius $r$. By the maximum modulus principle,

$\displaystyle |g(z)| = \frac{|f(z)|}{|z|} \leq \frac{|f(z_r)|}{|z_r|} \le \frac{1}{r}$

for all $z$ in $D_r$ and all $z_r$ on the boundary of $D_r$. As $r$ approaches $1$ we get $|g(z)| \leq 1$. Moreover, if there exists a $z_0$ in $D$ such that $g(z_0)=1$. Then, applying the maximum modulus principle to $g$, we obtain that $g$ is constant, hence $f(z)=kz$, where $k$ is constant and $|k|=1$. This is also the case if $|f'(0)|=1$.

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

Schwarz-Pick theorem: Let $f : D \to D$ be holomorphic. Then, for all $z_1, z_2 \in D$,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}$

and, for all $z \in D$

$\displaystyle\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}$.