update
This commit is contained in:
225
pages/Math416/Math416_L5.md
Normal file
225
pages/Math416/Math416_L5.md
Normal file
@@ -0,0 +1,225 @@
|
||||
# Lecture 5
|
||||
|
||||
## Review
|
||||
|
||||
Let $f$ be a complex function. that maps $\mathbb{R}^2$ to $\mathbb{R}^2$. $f(x+iy)=u(x,y)+iv(x,y)$.
|
||||
|
||||
$Df(x+iy)=\begin{pmatrix}
|
||||
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
|
||||
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
|
||||
\end{pmatrix}=\begin{pmatrix}
|
||||
\alpha & \beta\\
|
||||
\sigma & \delta
|
||||
\end{pmatrix}$
|
||||
|
||||
So
|
||||
|
||||
$$\begin{aligned}
|
||||
\frac{\partial f}{\partial \zeta}&=\frac{1}{2}\left(u_x+v_y\right)-i\frac{1}{2}\left(v_x+u_y\right)\\
|
||||
&=\frac{1}{2}\left(\alpha+\delta\right)-i\frac{1}{2}\left(\beta-\sigma\right)\\
|
||||
\end{aligned}$$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
\frac{\partial f}{\partial \overline{\zeta}}&=\frac{1}{2}\left(u_x+v_y\right)+i\frac{1}{2}\left(v_x+u_y\right)\\
|
||||
&=\frac{1}{2}\left(\alpha-\delta\right)+i\frac{1}{2}\left(\beta+\sigma\right)\\
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
When $f$ is conformal, $Df(x+iy)=\begin{pmatrix}
|
||||
\alpha & \beta\\
|
||||
-\beta & \alpha
|
||||
\end{pmatrix}$.
|
||||
|
||||
So $\frac{\partial f}{\partial \zeta}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a$
|
||||
|
||||
$\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}(\alpha-\alpha)+i\frac{1}{2}(\beta-\beta)=0$
|
||||
|
||||
> Less pain to represent a complex function using four real numbers.
|
||||
|
||||
## Chapter 3: Linear fractional Transformations
|
||||
|
||||
Let $a,b,c,d$ be complex numbers. such that $ad-bc\neq 0$.
|
||||
|
||||
The linear fractional transformation is defined as
|
||||
|
||||
$$
|
||||
\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}
|
||||
$$
|
||||
|
||||
If we let $\psi(\zeta)=\frac{e\zeta-f}{-g\zeta+h}$ also be a linear fractional transformation, then $\phi\circ\psi$ is also a linear fractional transformation.
|
||||
|
||||
New coefficients can be solved by
|
||||
|
||||
$$
|
||||
\begin{pmatrix}
|
||||
a & b\\
|
||||
c & d
|
||||
\end{pmatrix}
|
||||
\begin{pmatrix}
|
||||
e & f\\
|
||||
g & h
|
||||
\end{pmatrix}
|
||||
=
|
||||
\begin{pmatrix}
|
||||
k&l\\
|
||||
m&n
|
||||
\end{pmatrix}
|
||||
$$
|
||||
|
||||
So $\phi\circ\psi(\zeta)=\frac{k\zeta+l}{m\zeta+n}$
|
||||
|
||||
### Complex projective space
|
||||
|
||||
$\mathbb{R}P^1$ is the set of lines through the origin in $\mathbb{R}^2$.
|
||||
|
||||
We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{R}^2\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{R}\setminus\{0\}$ such that $(a,b)=t(c,d)$.
|
||||
|
||||
$R\mathbb{P}^1=S^1\setminus\{\pm x\}\cong S^1$
|
||||
|
||||
Equivalently,
|
||||
|
||||
$\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
|
||||
|
||||
We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{C}\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{C}\setminus\{0\}$ such that $(a,b)=(tc,td)$.
|
||||
|
||||
So, $\forall \zeta\in\mathbb{C}\setminus\{0\}$:
|
||||
|
||||
If $a\neq 0$, then $(a,b)\sim(1,\frac{b}{a})$.
|
||||
|
||||
If $a=0$, then $(0,b)\sim(0,-b)$.
|
||||
|
||||
So, $\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
|
||||
|
||||
### Linear fractional transformations
|
||||
|
||||
Let $M=\begin{pmatrix}
|
||||
a & b\\
|
||||
c & d
|
||||
\end{pmatrix}$ be a $2\times 2$ matrix with complex entries. That maps $\mathbb{C}^2$ to $\mathbb{C}^2$.
|
||||
|
||||
Suppose $M$ is non-singular. Then $ad-bc\neq 0$.
|
||||
|
||||
If $M\begin{pmatrix}
|
||||
\zeta_1\\
|
||||
\zeta_2
|
||||
\end{pmatrix}=\begin{pmatrix}
|
||||
\omega_1\\
|
||||
\omega_2
|
||||
\end{pmatrix}$, then $M\begin{pmatrix}
|
||||
t\zeta_1\\
|
||||
t\zeta_2
|
||||
\end{pmatrix}=\begin{pmatrix}
|
||||
t\omega_1\\
|
||||
t\omega_2
|
||||
\end{pmatrix}$.
|
||||
|
||||
So, $M$ induces a map $\phi_M:\mathbb{C}P^1\to\mathbb{C}P^1$ defined by $M\begin{pmatrix}
|
||||
\zeta\\
|
||||
1
|
||||
\end{pmatrix}=\begin{pmatrix}
|
||||
\frac{a\zeta+b}{c\zeta+d}\\
|
||||
1
|
||||
\end{pmatrix}$.
|
||||
|
||||
$\phi_M(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
|
||||
|
||||
If we let $M_2=\begin{pmatrix}
|
||||
e &f\\
|
||||
g &h
|
||||
\end{pmatrix}$, where $ad-bc\neq 0$ and $eh-fg\neq 0$, then $\phi_{M_2}(\zeta)=\frac{e\zeta+f}{g\zeta+h}$.
|
||||
|
||||
So, $M_2M_1=\begin{pmatrix}
|
||||
a&b\\
|
||||
c&d
|
||||
\end{pmatrix}\begin{pmatrix}
|
||||
e&f\\
|
||||
g&h
|
||||
\end{pmatrix}=\begin{pmatrix}
|
||||
\zeta\\
|
||||
1
|
||||
\end{pmatrix}$.
|
||||
|
||||
This also gives $\begin{pmatrix}
|
||||
k\zeta+l\\
|
||||
m\zeta+n
|
||||
\end{pmatrix}\sim\begin{pmatrix}
|
||||
\frac{k\zeta+l}{m\zeta+n}\\
|
||||
1
|
||||
\end{pmatrix}$.
|
||||
|
||||
So, if $ab-cd\neq 0$, then $\exists M^{-1}$ such that $M_2M_1=I$.
|
||||
|
||||
So non-constant linear fractional transformations form a group under composition.
|
||||
|
||||
When do two matrices gives the $t_0$ same linear fractional transformation?
|
||||
|
||||
$M_2^{-1}M_1=\alpha I$
|
||||
|
||||
We defined $GL(2,\mathbb{C})$ to be the group of general linear transformations of order 2 over $\mathbb{C}$.
|
||||
|
||||
This is equivalent to the group of invertible $2\times 2$ matrices over $\mathbb{C}$ under matrix multiplication.
|
||||
|
||||
Let $F$ be the function that maps $M$ to $\phi_M$.
|
||||
|
||||
$F:GL(2,\mathbb{C})\to\text{Homeo}(\mathbb{C}P^1)$
|
||||
|
||||
So the kernel of $F$ is the set of matrices that represent the identity transformation. $\ker F=\left\{\alpha I\right\},\alpha\in\mathbb{C}\setminus\{0\}$.
|
||||
|
||||
#### Corollary of conformality
|
||||
|
||||
If $\phi$ is a non-constant linear fractional transformation, then $\phi$ is conformal.
|
||||
|
||||
Proof:
|
||||
|
||||
Know that $\phi_0\circ\phi(\zeta)=\zeta$,
|
||||
|
||||
Then $\phi(\zeta)=\phi_0^{-1}\circ\phi\circ\phi_0(\zeta)$.
|
||||
|
||||
So $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
|
||||
|
||||
$\phi:\mathbb{C}\cup\{\infty\}\to\mathbb{C}\cup\{\infty\}$ which gives $\phi(\infty)=\frac{a}{c}$ and $\phi(-\frac{d}{c})=\infty$.
|
||||
|
||||
So, $\phi$ is conformal.
|
||||
|
||||
EOP
|
||||
|
||||
#### Proposition 3.4 of Fixed points
|
||||
|
||||
Any non-constant linear fractional transformation except the identity transformation has 1 or 2 fixed points.
|
||||
|
||||
Proof:
|
||||
|
||||
Let $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
|
||||
|
||||
Case 1: $c=0$
|
||||
|
||||
Then $\infty$ is a fixed point.
|
||||
|
||||
Case 2: $c\neq 0$
|
||||
|
||||
Then $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
|
||||
|
||||
The solution of $\phi(\zeta)=\zeta$ is $c\zeta^2+(d-a)\zeta-b=0$.
|
||||
|
||||
Such solutions are $\zeta=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
|
||||
|
||||
So, $\phi$ has 1 or 2 fixed points.
|
||||
|
||||
EOP
|
||||
|
||||
#### Proposition 3.5 of triple transitivity
|
||||
|
||||
If $\zeta_1,\zeta_2,\zeta_3\in\mathbb{C}P^1$ are distinct, then there exists a non-constant linear fractional transformation $\phi$ such that $\phi(\zeta_1)=\zeta_2$ and $\phi(\zeta_3)=\infty$.
|
||||
|
||||
Proof as homework.
|
||||
|
||||
#### Theorem 3.8 Preservation of clircles
|
||||
|
||||
We defined clircle to be a circle or a line.
|
||||
|
||||
If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
|
||||
|
||||
Proof:
|
||||
|
||||
Continue on next lecture.
|
||||
@@ -7,4 +7,5 @@ export default {
|
||||
Math416_L2: "Complex Variables (Lecture 2)",
|
||||
Math416_L3: "Complex Variables (Lecture 3)",
|
||||
Math416_L4: "Complex Variables (Lecture 4)",
|
||||
Math416_L5: "Complex Variables (Lecture 5)",
|
||||
}
|
||||
|
||||
Reference in New Issue
Block a user