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how does one find the set of Automorphisms of the complex projective line?

PS: no scheme theory is assumed.

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    This link may be useful: https://en.wikipedia.org/wiki/M%C3%B6bius_transformation2012-09-03

1 Answers 1

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0) I'll use coordinates $(t:z)$ on the projective line $\mathbb P^1(\mathbb C)$, with the embedding $\mathbb C\to \mathbb P^1(\mathbb C)$ given by $z\mapsto (1:z)$ and with $\infty = (0:1)$.

1) Now, given an automorphism $f:\mathbb P^1(\mathbb C)\to \mathbb P^1(\mathbb C)$, we may assume $f(\infty)=\infty$: if this is not the case and if $f(\infty)=a=(1:a)\in \mathbb C$, we consider the automorphism $g:\mathbb P^1(\mathbb C)\to \mathbb P^1(\mathbb C): (t:z)\mapsto (z-at:t)$ (which maps $a=(1:a)$ to $(0:1)=\infty$) and the new automorphism $g\circ f$ will satisfy $(g\circ f)(\infty)=\infty$.

2) If $f(\infty)=\infty$, we may consider the restriction $f\mid \mathbb C=f_0:\mathbb C\to\mathbb C$.
It is given by a polynomial $f_0(z)=P(z)$ and since it is a bijection it must have degree $1$ (by the fundamental theorem of algebra, say) : $P(z)=bz+c=(1:bz+c)$ .

3) Taking into account the reduction in 1), we see that the original automorphism must have the form $(t:z) \mapsto (\gamma z+\delta t:\alpha z+\beta t) $ with $\alpha,\beta,\gamma,\delta \in \mathbb C$ and $\alpha\delta-\beta\gamma\neq0$.
Conversely, it is clear that such a formula defines an automorphism of $\mathbb P^1(\mathbb C)$.
With the obvious traditional abuse of notation we just write this as the Möbius transformation $f(z)=\frac {\alpha z+\beta}{\gamma z+\delta} $

5) Summary $Aut(P^1(\mathbb C)) =PGl_2(\mathbb C)=Gl_2(\mathbb C)/ \mathbb C ^* $

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    Dear @Gerard, I agree with you that from the purely gemetric point of view inversion and conjugation on the Riemann sphere are quite similar, but the fact remains that the first is holomorphic and the second is not. *C'est la vie...*2013-04-28