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Define a map from $2 \times 2$ invertible matrices to linear fractional transformations $$ f:\left( \begin{array}{ccc} a & b \\ c & d \\\end{array} \right) \mapsto \frac{az + b}{cz + d}.$$ It is well known that $f(AB) = f(A) \circ f(B)$. This is easy to prove: just simplify both sides of the equation. But is there a more conceptual proof that does not involve this computation? Is there a generalization to $3 \times3$ matrices, etc.?

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    $\begingroup$ Yes. See here for a description of homogeneous coordinates on the Riemann sphere. The natural generalization is the action of any projective linear group on the corresponding projective space. $\endgroup$ – Jim Belk Nov 12 '13 at 4:26
  • $\begingroup$ Ah, I see. Thanks for the links! $\endgroup$ – Jair Taylor Nov 13 '13 at 18:39
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The complex general linear group ${\rm GL}_2(\Bbb C)$ acts on $\Bbb C^2$. Since matrices act linearly, there is an induced action on complex projective space (the collection of one-dimensional subspaces).

This space ${\Bbb P}^1(\Bbb C)$ can be defined by putting an equivalence relation on the set of nonzero points $(z,w)$ of $\Bbb C^2$, where $(\lambda z,\lambda w)\sim(z,w)$ for all $\lambda\in\Bbb C^\times$: then the equivalence classes will be the one-dimensional subspaces of $\Bbb C^2$ (not counting the origin, since all subspaces contain $0$). All points can be represented as either a tuple $(z,1)$ with $z\in\Bbb C$ arbitrary or as $(1,0)$. If we think of $(z,w)$ as the ratio $z/w$ on the so-called Riemann sphere $\widehat{\Bbb C}$, the point $(z,1)$ represents $z\in\Bbb C$ while $(1,0)$ represents $\infty$, according to the convention $1/0=\infty$.

In summary, the action of ${\rm GL}_2(\Bbb C)$ on $\Bbb C^2$ induces an action on ${\Bbb P}^1(\Bbb C)$ which can be transported to an action on the Riemann sphere $\widehat{\Bbb C}=\Bbb C\cup\{\infty\}$. Writing out the action explicitly yields these linear fractional transformations, also called Mobius transformations: $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]z=\frac{az+b}{cz+d}$.

If a group $G$ acts on a space and $H$ is the subgroup of elements acting as the identity map, then we can descend to an action of $G/H$ on the space. Since the subgroup of multiplication-by-scalars in the general linear group maps each line to itself in ${\Bbb P}^1(\Bbb C)$, after collapsing superfluous redundancy we "really" have an action of ${\rm PSL}_2(\Bbb C)$ on ${\Bbb P}^1(\Bbb C)$.

More generally, an action of ${\rm GL}_n(\Bbb C)$ descends to an action of ${\rm PGL}_n(\Bbb C)$ on ${\Bbb P}^{n-1}(\Bbb C)$.

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  • $\begingroup$ Hmm, I don't think I saw this the first time around. Nice answer - not too late for a +1 and accept! $\endgroup$ – Jair Taylor Apr 14 '16 at 18:13

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