## 数学代写|复分析作业代写Complex function代考|The Cross-Ratio

Returning to (3.25), we have already established that if we can find a Möbius transformation $M$ that maps three given points $q, r, s$ to three other given points $\widetilde{q}, \widetilde{r}, \widetilde{s}$, then $M$ is unique. It thus remains to show that such an $M$ always exists.

To see this, first let us arbitrarily choose three points $q^{\prime}, r^{\prime}, s^{\prime}$, once and for all. Next, suppose we can write down a Möbius transformation mapping three arbitrary points $q, r, s$ to these particular three points, $q^{\prime}, r^{\prime}, s^{\prime} ;$ let $M_{q r s}(z)$ denote this Möbius transformation. In exactly the same way we could also write down $M_{\bar{q} \bar{r} \bar{s}}(z)$. By virtue of the group property, it is now easy to see that
$$M=M_{\tilde{q} \tilde{r} \tilde{s}}^{-1} \circ M_{\mathrm{qrs}}$$ is a Möbius transformation mapping $q, r, s$ to $q^{\prime}, r^{\prime}, s^{\prime}$ and thence to $\widetilde{q}, \widetilde{r}, \widetilde{s}$, as was desired.

Now the real trick is to choose $q^{\prime}, r^{\prime}, s^{\prime}$ in such a way as to make it easy to write down $M_{\mathrm{qrs}}(z)$. We don’t like to pull rabbits out of hats, but try $\mathrm{q}^{\prime}=0, r^{\prime}=1$, and $s^{\prime}=\infty$. Along with this special choice comes a special, standard notation: the is written $[z, \mathrm{q}, \mathrm{r}, \mathrm{s}]$.

In order to map $q$ to $q^{\prime}=0$ and $s$ to $s^{\prime}=\infty$, the numerator and denominator of $[z, q, r, s]$ must be proportional to $(z-q)$ and $(z-s)$, respectively. Thus $[z, q, r, s]=$ $k\left(\frac{z-q}{z-s}\right)$, where $k$ is a constant. Finally, since $k\left(\frac{r-q}{r-s}\right)=[r, q, r, s] \equiv 1$, we deduce that
$$[z, q, r, s]=\frac{(z-q)(r-s)}{(z-s)(r-q)}$$

## 数学代写|复分析作业代写Complex function代考|Empirical Evidence of a Link with Linear Algebra

As you were reading about the group property of Möbius transformations, you may well have experienced déjà $v u$, for the results we obtained were remarkably reminiscent of the behaviour of matrices in linear algebra. Before explaining the reason for this connection between Möbius transformations and linear algebra, let us be more explicit about the empirical evidence for believing that such a connection exists.

We begin by associating with every Möbius transformation $M(z)$ a corresponding $2 \times 2$ matrix $[\mathrm{M}]$ :
$$M(z)=\frac{a z+b}{c z+d} \quad \longleftrightarrow \quad[M]=\left[\begin{array}{ll} a & b \ c & d \end{array}\right]$$
Since the coefficients of the Möbius transformation are not unique, neither is the corresponding matrix: if $k$ is any non-zero constant, then the matrix $k[M]$ corresponds to the same Möbius transformation as $[M]$. However, if $[M]$ is normalized by imposing $(\mathrm{ad}-\mathrm{bc})=1$, then there are just two possible matrices associated with a given Möbius transformation: if one is called $[M]$, the other is $-[M]$; in other words, the matrix is determined “uniquely up to sign”. This apparently trivial fact turns out to have deep significance in both mathematics and physics; see Penrose and Rindler (1984, Ch. 1) and Penrose (2005).

At this point there exists a strong possibility of confusion, so we issue the following WARNING: In linear algebra we are-or should be!-accustomed to thinking of a real $2 \times 2$ matrix as representing a linear transformation of $\mathbb{R}^2$. For example, $\left(\begin{array}{cc}0 & -1 \ 1 & 0\end{array}\right)$ represents a rotation of the plane through $(\pi / 2)$. That is, when we apply it to a vector $\left(\begin{array}{l}x \ y\end{array}\right)$ in $\mathbb{R}^2$, we obtain
$$\left(\begin{array}{rr} 0 & -1 \ 1 & 0 \end{array}\right)\left(\begin{array}{l} x \ y \end{array}\right)=\left(\begin{array}{r} -y \ x \end{array}\right)=\left{\left(\begin{array}{l} x \ y \end{array}\right) \text { rotated by }(\pi / 2)\right} .$$
In stark contrast, the matrix $\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$ corresponding to a Möbius transformation generally has complex numbers as its entries, and so it cannot be interpreted as a linear transformation of $\mathbb{R}^2$. Even if the entries are real, it must not be thought of in this way. For example, the matrix $\left(\begin{array}{cc}0 & -1 \ 1 & 0\end{array}\right)$ corresponds to the Möbius transformation $M(z)=-(1 / z)$, which is certainly not a linear transformation of $\mathbb{C}$. To avoid confusion, we will adopt the following notational convention: We use (ROUND) brackets for a real matrix corresponding to a linear transformation of $\mathbb{R}^2$ or of $\mathbb{C}$, and we use [SQUARE] brackets for a (generally) complex matrix corresponding to a Möbius transformation of $\mathbb{C}$.

# 复分析代考

## 数学代写|复分析作业代写Complex function代考|The Cross-Ratio

$$M=M_{\tilde{q} \tilde{r} \tilde{s}}^{-1} \circ M_{\mathrm{qrs}}$$

$$[z, q, r, s]=\frac{(z-q)(r-s)}{(z-s)(r-q)}$$

## 数学代写|复分析作业代写Complex function代考|Empirical Evidence of a Link with Linear Algebra

$$M(z)=\frac{a z+b}{c z+d} \quad \longleftrightarrow \quad[M]=\left[\begin{array}{lll} a & b c & d \end{array}\right]$$

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