NoteNextra-origin/content/Math416/Math416_L5.md

Math416 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 z}&=\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{z}}&=\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 z}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a

\frac{\partial f}{\partial \overline{z}}=\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(z)=\frac{az+b}{cz+d}

If we let \psi(z)=\frac{ez-f}{-gz+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(z)=\frac{kz+l}{mz+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 z\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 $M\begin{pmatrix} z_1\ z_2 \end{pmatrix}=\begin{pmatrix} \omega_1\ \omega_2 \end{pmatrix}$, then $M\begin{pmatrix} tz_1\ tz_2 \end{pmatrix}=\begin{pmatrix} t\omega_1\ t\omega_2 \end{pmatrix}$.

So, M induces a map \phi_M:\mathbb{C}P^1\to\mathbb{C}P^1 defined by $M\begin{pmatrix} z\ 1 \end{pmatrix}=\begin{pmatrix} \frac{az+b}{cz+d}\ 1 \end{pmatrix}$.

\phi_M(z)=\frac{az+b}{cz+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}(z)=\frac{ez+f}{gz+h}.

So, $M_2M_1=\begin{pmatrix} a&b\ c&d \end{pmatrix}\begin{pmatrix} e&f\ g&h \end{pmatrix}=\begin{pmatrix} z\ 1 \end{pmatrix}$.

This also gives $\begin{pmatrix} kz+l\ mz+n \end{pmatrix}\sim\begin{pmatrix} \frac{kz+l}{mz+n}\ 1 \end{pmatrix}$.

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(z)=z,

Then \phi(z)=\phi_0^{-1}\circ\phi\circ\phi_0(z).

So \phi(z)=\frac{az+b}{cz+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.

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(z)=\frac{az+b}{cz+d}.

Case 1: c=0

Then \infty is a fixed point.

Case 2: c\neq 0

Then \phi(z)=\frac{az+b}{cz+d}.

The solution of \phi(z)=z is cz^2+(d-a)z-b=0.

Such solutions are z=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}.

So, \phi has 1 or 2 fixed points.

Proposition 3.5 of triple transitivity

If z_1,z_2,z_3\in\mathbb{C}P^1 are distinct, then there exists a non-constant linear fractional transformation \phi such that \phi(z_1)=z_2 and \phi(z_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.