I saw on my complex analysis book that linear fractional transformation is isomorphic to the group of invertable $2\times 2$ matrix such that identify scalar multiplication.

Verifying that was easy but I want to know whether there is some intuition or underlying principles why this is happening. I was curious about it since high school. (at that time it was about real fractional transformation)

In fact I didn't take modern algebra and I do not know much about group things.(Just learn a little when taking topology class). Can someone explain the reasons?


3 Answers 3


$(1)$ Let $A$ be any matrix in $M_2(\Bbb C)$, set $h_A(z)=\dfrac{az+b}{cz+d}$. One can check that $h'_A(z)=\dfrac{\det A}{(cz+d)^2}$ which motivates taking $A\in {\rm GL}(2,\Bbb C)$. Now, if we multiply $A$ by any scalar, i.e $B=\alpha A$, then $h_A=h_B$ since in$\frac{\alpha az+\alpha b}{\alpha cz+\alpha d}$ the $\alpha$s cancel out. In particular $h_A=h_{\rm id}$ means that $A$ is a multiple of the identity, so the map $A\mapsto h_A$ has kernel $\alpha I$, and we get that the group $G$ of Möbius transformations is isomorphic to ${\rm GL}(2,\Bbb C)/\{\alpha\cdot 1\}={\rm PGL}(2,\Bbb C)$.

$(2)$ We can do something more. Since $\det A\neq 0$; we can assume that $\det A=1$ by taking $\alpha =(\det A)^{-1}$. Thus we can take $A\in {\rm SL}(2,\Bbb C)$. And consider the still surjective map $A\in {\rm SL}(2,\Bbb C)\mapsto h_A$. You can check this has kernel $\{1,-1\}$ so that $G$ is isomorphic to ${\rm SL}(2,\Bbb C)/\{\pm 1\}={\rm PSL}(2,\Bbb C)$, too.

As you can see, there isn't much mystery to this, really.

  • $\begingroup$ Does $PGL(2,\Bbb C)\cong PSL(2,\Bbb C)$? $\endgroup$
    – Ooker
    Nov 13, 2017 at 18:14
  • $\begingroup$ @Ooker Yes, by what I wrote above. $\endgroup$
    – Pedro
    Nov 13, 2017 at 18:27
  • 1
    $\begingroup$ @Ooker In fact there is a canonical isomorphism $PGL(n, \Bbb F) / PSL(n, \Bbb F) \cong \Bbb F^* / (\Bbb F^*)^n$, so $PGL(n, \Bbb F) \cong PSL(n, \Bbb F)$ iff every element of $\Bbb F^*$ has an $n$th root. In particular, this is true for algebraically closed $\Bbb F$ (like $\Bbb C$) for every $n$. By contrast, positive elements of $\Bbb R$ admit square roots but negative elements do not, so $[PGL(2, \Bbb R) : PSL(2, \Bbb R)] = 2$. $\endgroup$ Jul 5, 2022 at 10:17

To address the question of intuition behind this isomorphism: mobius transformations act on the Riemann sphere $\hat{\mathbb{C}}$ by dilations, rotations, and reflections. The Riemann sphere can be identified with $\mathbb{P}^1(\mathbb{C})$, and it is under this identification that mobius transformations become elements of $PGL(2, \mathbb{C})$.

Edit: I can explain my answer better.

  1. "Mobius transformations act on the Riemann sphere by dilations, rotations, and reflections." Ok, this I do have a reference for: Gamelin's "Complex Analysis" proves that any fractional linear transformation (which is the same this as a mobius transformation) is the composition of dilations, translations, and inversions.
  2. "The Riemann Sphere can be identified with $\mathbb{P}^1(\mathbb{C})$." Recall that $\mathbb{P}^1(\mathbb{C})$ is the set of all complex lines in $\mathbb{C}^2$, eg. $az_1 + bz_2 = 0$. Notice that this line is almost determined by $a$ and $b$; the only problem is that if we multiply them both by any non-zero complex number, say $\lambda$, the line remains the same. (Zeros of $az_1+bz_2$ will also be zeros of $ \lambda az_1 + \lambda b z_2$.) We call $[a:b]$ homogeneous coordinates, and agree that $[a:b] = [\lambda a : \lambda b]$ for any non-zero $\lambda \in \mathbb{C}$. For examples, provided $a\neq 0$, we can write $[a:b] = [1:b/a]$ for any $b\in \mathbb{C}$, by taking $\lambda = \frac{1}{a}$. Exercise: show that the map $f: \hat{\mathbb{C}} \to\mathbb{P}^1(\mathbb{C})$ that sends $\infty$ to $[0:1]$ and $\infty\neq z\in \hat{\mathbb{C}}$ to $[1:z]$ is well-defined, continuous, invertible, etc.

  3. "It is under this identification that mobius transformations become elements of $PGL(2, \mathbb{C})$." An element of $PGL(2, \mathbb{C})$ is a 2-by-2 matrix with complex coefficients, up to multiplication by a scalar matrix, i.e. diagonal matrix of the form $\lambda\cdot \hbox{Id}$. So for example, in $PGL(2, \mathbb{C})$, $$\big(\begin{matrix} 1 & i\\0 & 2\\ \end{matrix}\big) = \big(\begin{matrix} i & -1\\0 & 2i\\ \end{matrix}\big)$$ since the second matrix above is the first one times the matrix $\big(\begin{matrix} i & 0\\0 & i\\ \end{matrix}\big)$. Some questions to ask yourself if this group doesn't seem natural:

    i) If $\mathbb{P}^1(\mathbb{C})$ is the set of lines in $\mathbb{C}^2$, what elements of $GL(2, \mathbb{C})$ send each line to itself?

    ii) Can we find an action of $GL(2, \mathbb{C})$ on $\mathbb{P}^1(\mathbb{C})$? (Hint: use coordinates)

    iii) Since some elements of $GL(2, \mathbb{C}$ fix every element of $\mathbb{P}^1(\mathbb{C})$, is there any harm in quotienting out by this group?

    Once you are comfortable with $PGL(2, \mathbb{C})$, do the following: grab an element of this group; lift it to a genuine matrix (that is, fix a specific matrix representation of it); observe its action on (0, 1) and ( 1, $z$); use #2 to interpret this on $\hat{\mathbb{C}}$

That ought to do it.

  • $\begingroup$ Thank you. Can you recommend a book or a pdf file that explains about you answered? $\endgroup$ Dec 11, 2016 at 9:47

What is your definition of $PGL_n(\mathbb{C})$ ? To me it is the set of transformations of the form $$[x_1,\ldots,x_n,1] \mapsto [ (M x)_1,\ldots, (M x)_n,(M x)_{n+1}] =\left[ \frac{(M x)_1}{(M x)_{n+1}},\ldots, \frac{(M x)_n}{(M x)_{n+1}},1\right]$$ where $M \in GL_{n+1}(\mathbb{C})$ and $x_{n+1} = 1$.

With $n=1$ it is $x \mapsto \frac{ax+b}{cx+d}$ with $ad-bc \ne 0$, the set of Möbius transformations.

  • $\begingroup$ Standard definition: en.wikipedia.org/wiki/Projective_linear_group $\endgroup$ Jan 4, 2017 at 18:08
  • $\begingroup$ @DavidSteinberg The 1st sentence is $PGL_n(k)$ is the induced action of the general linear group of $k^{n+1}$ on the associated projective space $P^n(k)$, i.e. what I wrote. $\endgroup$
    – reuns
    Jan 4, 2017 at 18:16

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