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I know that Möbius transformations generally maps lines/circles to line/circles using a function $f\left( z \right) = \frac{{az + b}}{{cz + d}}$ defined over $\mathbb C$. However, what I do not understand is that they usually are represented via a matrix and often written as

$$\left[ {\begin{array}{*{20}{c}} z&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&c\\ b&d \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {az + b}&{cz + d} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{az + b}}{{cz + d}}}&1 \end{array}} \right] = f\left( z \right)$$

I took the last equation from this Wikipedia article. I cannot understand the meaning of this equation and how could you equate $f(z)$ to the row vector $\left[ {\begin{array}{*{20}{c}} {\frac{{az + b}}{{cz + d}}}&1 \end{array}} \right]$. Could someone put these in simple terms so that I see how things work. Thanks in advance.

user829347
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Alex Mathy
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1 Answers1

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Geometrically, the set of lines through the origin in $\mathbf{C}^{2}$ corresponds to an equivalence relation on non-zero vectors: $$ (Z_{0}, Z_{1}) \sim (tZ_{0}, tZ_{1}),\quad t \neq 0. $$ Particularly, if $Z_{1} \neq 0$ and we define $z_{0} = Z_{0}/Z_{1}$, then $(Z_{0}, Z_{1}) \sim (Z_{0}/Z_{1}, 1) = (z_{0}, 1)$. Similarly, if $Z_{0} \neq 0$ and we define $z_{1} = Z_{1}/Z_{0}$, then $(Z_{0}, Z_{1}) \sim (1, Z_{1}/Z_{0}) = (1, z_{1})$. (The diagram comes from a question about visualizing the projective line. Over the complex numbers, the projective line is identified with the Riemann sphere.)

Lines through the origin in the plane

If $ad - bc \neq 0$, multiplication by the matrix $\left[\begin{array}{@{}cc@{}}a & c \\ b & d \\ \end{array}\right]$ induces a bijection on the set of non-zero vectors, and maps lines through the origin to lines through the origin. Via the interpretation of the preceding paragraph, this matrix induces a (holomorphic) bijection of the Riemann sphere to itself. Finally to answer the question, if $cz + d \neq 0$, then $$ (az + b, cz + d) \sim \Bigl(\frac{az + b}{cz + d}, 1\Bigr). $$

Andrew D. Hwang
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  • what does it mean for two vectors to be equivalent? is it the equivalence of two lines to be parallel on the set of reals $\mathbb R$? – Alex Mathy Jan 11 '22 at 16:04
  • Two vectors are equivalent in this sense if any only if they are mutually-proportional, i.e., they determine the same line through $(0, 0)$ in $\mathbf{C}^{2}$. (Incidentally, it may be a bit more common to transpose everything, writing vectors as columns on the right, but that's merely a notational difference.) – Andrew D. Hwang Jan 11 '22 at 17:27