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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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    What's your definition of a projective variety and of a morphism of projective varieties?2012-09-03
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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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    Awesome explanation, thanks.2012-09-04
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    You are welcome, GjR.2012-09-04
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    Why is it that $f(z)=\overline{z}$ does not belong to the automorphisms? Seen from the point of view of the Riemann sphere there is e.g. no difference between complex conjugation or an inversion like $f(z)= 1/z$.2013-04-28
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    @Gerard: I have classified the *holomorphic* automorphisms of $\mathbb P^1(\mathbb C)$. Of course there are tons of non-holomorphic diffeomorphisms too, and $z\mapsto \bar z$ is one of them. Contrary to what you write inversion is very different from conjugation: the former is holomorphic, the latter is not.2013-04-28
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    I have to look up the difference between holomorphic/non-holomorphic, automorphism/diffeomorphism. On the Riemann sphere the inversion is a reflection in a horizontal plane through the center of the sphere, complex conjugation a reflection in a vertical plane through the center of the sphere, I am inclined to think that such reflections resemble each other (from the point of view of this sphere).2013-04-28
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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